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Prove that the following are true for all positive integral value of n. Use mathematical induction.  2.

Pertanyaan

Prove that the following are true for all positive integral value of n. Use mathematical induction. 

2.  straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight n minus 1 end exponent equals fraction numerator straight a open parentheses 1 minus straight r to the power of straight n close parentheses over denominator 1 minus straight r end fraction space  

Pembahasan Soal:

Membuktikan dengan cara induksi matematika dimana untuk nilai n =1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight n minus 1 end exponent end cell equals cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight n close parentheses over denominator 1 minus straight r end fraction space end cell row cell straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of 1 minus 1 end exponent end cell equals cell fraction numerator straight a open parentheses 1 minus straight r to the power of 1 close parentheses over denominator 1 minus straight r end fraction end cell row cell straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of 0 end cell equals cell fraction numerator straight a open parentheses 1 minus straight r to the power of 1 close parentheses over denominator 1 minus straight r end fraction end cell row straight a equals cell straight a rightwards arrow terbukti end cell end table

Untuk n = k diasumsikan terbukti maka

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight n minus 1 end exponent end cell equals cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight n close parentheses over denominator 1 minus straight r end fraction space end cell row cell straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of k minus 1 end exponent end cell equals cell fraction numerator straight a open parentheses 1 minus straight r to the power of k close parentheses over denominator 1 minus straight r end fraction rightwards arrow terbukti end cell end table

Untuk n = k+1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight n minus 1 end exponent end cell equals cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight n close parentheses over denominator 1 minus straight r end fraction space end cell row cell straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight k minus 1 end exponent plus straight a. straight r to the power of straight k plus 1 minus 1 end exponent end cell equals cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction end cell row cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight k close parentheses over denominator 1 minus straight r end fraction plus ar to the power of straight k end cell equals cell space fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction end cell row cell fraction numerator straight a minus ar to the power of straight k plus left parenthesis 1 minus straight r right parenthesis ar to the power of straight k over denominator 1 minus straight r end fraction end cell equals cell space fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction end cell row cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction end cell equals cell space fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction rightwards arrow terbukti end cell end table

Jadi terbukti bahwa straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight n minus 1 end exponent equals fraction numerator straight a open parentheses 1 minus straight r to the power of straight n close parentheses over denominator 1 minus straight r end fraction space karena hasil sisi kiri dan kanan sama

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Mahasiswa/Alumni UIN Walisongo Semarang

Terakhir diupdate 11 Juli 2021

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Pertanyaan yang serupa

Diberikan pernyataan   untuk setiap bilangan asli . Dengan menggunakan induksi matematika, dapat disimpulkan bahwa ....

Pembahasan Soal:

Dimisalkan pernyataan P subscript n sebagai berikut.

 P subscript n colon space 1 plus 2 plus 3 plus horizontal ellipsis plus n equals open parentheses n plus begin display style 1 half end style close parentheses squared over 2

untuk setiap bilangan asli n.

Karena akan dibuktikan pernyataan untuk setiap bilangan asli n, yaitu n greater or equal than 1, langkah pertamanya adalah buktikan P subscript 1 benar.


LANGKAH 1: Buktikan bold italic P subscript bold 1 benar.

Perhatikan pernyataan P subscript n sebagai berikut!

 

 P subscript n colon space 1 plus 2 plus 3 plus horizontal ellipsis plus n equals open parentheses n plus begin display style 1 half end style close parentheses squared over 2

untuk setiap bilangan asli n.

Oleh karena itu, pernyataan P subscript 1 bisa didapat dengan melakukan substitusi n equals 1 ke dalam pernyataan P subscript n sebagai berikut.

P subscript 1 colon space 1 equals open parentheses 1 plus begin display style 1 half end style close parentheses squared over 2

Ruas kiri pada pernyataan tersebut adalah 1.

Kemudian, ruas kanan pada pernyataan tersebut dapat disederhanakan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 plus begin display style 1 half end style close parentheses squared over 2 end cell equals cell open parentheses begin display style 3 over 2 end style close parentheses squared over 2 end cell row blank equals cell fraction numerator begin display style 9 over 4 end style over denominator 2 end fraction end cell row blank equals cell 9 over 4 divided by 2 end cell row blank equals cell 9 over 4 cross times 1 half end cell row blank equals cell 9 over 8 end cell end table

Dapat diperhatikan bahwa nilai di ruas kiri berbeda dengan nilai di ruas kanan. Oleh karena itu, P subscript 1 bernilai SALAH.

Dengan demikian, terdapat kesalahan pada langkah pertama.

 

LANGKAH 2: Buktikan untuk sembarang bilangan asli bold italic k, jika bold italic P subscript bold italic k bernilai benar mengakibatkan bold italic P subscript bold italic k bold plus bold 1 end subscript bernilai benar.

Perhatikan pernyataan P subscript n sebagai berikut!

 P subscript n colon space 1 plus 2 plus 3 plus horizontal ellipsis plus n equals open parentheses n plus begin display style 1 half end style close parentheses squared over 2

untuk setiap bilangan asli n.

Asumsikan pernyataan P subscript k bernilai BENAR.

Pernyataan P subscript k bisa didapat dengan melakukan substitusi n equals k ke dalam pernyataan P subscript n sebagai berikut.

P subscript k colon space 1 plus 2 plus 3 plus horizontal ellipsis plus k equals open parentheses k plus begin display style 1 half end style close parentheses squared over 2

Selanjutnya, akan dicek nilai kebenaran dari pernyataan P subscript k plus 1 end subscript.

Pernyataan P subscript k plus 1 end subscript bisa didapat dengan melakukan substitusi n equals k plus 1 ke dalam pernyataan P subscript n sebagai berikut.

P subscript k plus 1 end subscript colon space 1 plus 2 plus 3 plus horizontal ellipsis plus k plus open parentheses k plus 1 close parentheses equals open parentheses open parentheses k plus 1 close parentheses plus begin display style 1 half end style close parentheses squared over 2

Pada ruas kiri P subscript k plus 1 end subscript, didapat perhitungan sebagai berikut.

1 plus 2 plus 3 plus horizontal ellipsis plus k plus open parentheses k plus 1 close parentheses equals open parentheses k plus begin display style 1 half end style close parentheses squared over 2 plus open parentheses k plus 1 close parentheses equals fraction numerator k squared plus k plus begin display style 1 fourth end style over denominator 2 end fraction plus fraction numerator 2 k plus 2 over denominator 2 end fraction equals fraction numerator k squared plus 3 k plus begin display style 9 over 4 end style over denominator 2 end fraction equals open parentheses k plus begin display style 3 over 2 end style close parentheses squared over 2 equals open parentheses open parentheses k plus 1 close parentheses plus begin display style 1 half end style close parentheses squared over 2

Dari penjabaran di atas, didapatkan bahwa bentuk pada ruas kiri P subscript k plus 1 end subscript sama dengan bentuk pada ruas kanannya sehingga P subscript k plus 1 end subscript bernilai benar.

Dengan demikian, pada proses pembuktian dengan induksi matematika, disimpulkan bahwa pernyataan tidak terbukti karena terdapat kesalahan pada langkah pertama.

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Buktikan bahwa:  a.

Pembahasan Soal:

Membuktikan dengan induksi matematika dimana

Untuk n = 1

table attributes columnalign right center left columnspacing 0px end attributes row cell straight P subscript straight n end cell identical to cell fraction numerator 1 over denominator 1 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 5 end fraction plus... plus fraction numerator 1 over denominator open parentheses 2 straight n minus 1 close parentheses open parentheses 2 straight n plus 1 close parentheses end fraction equals fraction numerator straight n over denominator 2 straight n plus 1 end fraction end cell row blank blank cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 1 third equals 1 third rightwards arrow terbukti end cell row blank blank blank end table

Untuk n = k diasumsikan terbukti maka

table attributes columnalign right center left columnspacing 0px end attributes row cell straight P subscript straight n end cell identical to cell fraction numerator 1 over denominator 1 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 5 end fraction plus... plus fraction numerator 1 over denominator open parentheses 2 straight n minus 1 close parentheses open parentheses 2 straight n plus 1 close parentheses end fraction equals fraction numerator straight n over denominator 2 straight n plus 1 end fraction end cell row cell straight P subscript straight k end cell identical to cell fraction numerator 1 over denominator 1 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 5 end fraction plus... plus fraction numerator 1 over denominator open parentheses 2. straight k minus 1 close parentheses open parentheses 2. straight k plus 1 close parentheses end fraction equals fraction numerator straight k over denominator 2. straight k plus 1 end fraction end cell row blank blank blank end table

Untuk n = k+1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell straight P subscript straight n end cell identical to cell fraction numerator 1 over denominator 1 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 5 end fraction plus... plus fraction numerator 1 over denominator open parentheses 2 straight n minus 1 close parentheses open parentheses 2 straight n plus 1 close parentheses end fraction equals fraction numerator straight n over denominator 2 straight n plus 1 end fraction end cell row cell straight P subscript straight k plus 1 end subscript end cell identical to cell fraction numerator 1 over denominator 1 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 5 end fraction plus... plus fraction numerator 1 over denominator open parentheses 2 straight k minus 1 close parentheses open parentheses 2 straight k plus 1 close parentheses end fraction plus fraction numerator 1 over denominator open parentheses 2 straight k plus 1 close parentheses open parentheses 2 straight k plus 3 close parentheses end fraction equals fraction numerator straight k plus 1 over denominator 2 straight k plus 3 end fraction rightwards arrow terbukti end cell row blank blank blank end table

Akan dibuktikan bahwa

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator 1 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 5 end fraction plus... plus fraction numerator 1 over denominator open parentheses 2 straight n minus 1 close parentheses open parentheses 2 straight n plus 1 close parentheses end fraction end cell equals cell fraction numerator straight n over denominator 2 straight n plus 1 end fraction end cell row cell fraction numerator 1 over denominator 1 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 5 end fraction plus... plus fraction numerator 1 over denominator open parentheses 2 straight k minus 1 close parentheses open parentheses 2 straight k plus 1 close parentheses end fraction plus fraction numerator 1 over denominator open parentheses 2 straight k plus 1 close parentheses open parentheses 2 straight k plus 3 close parentheses end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 2 straight k plus 3 end fraction end cell row cell fraction numerator k over denominator 2 straight k plus 1 end fraction plus fraction numerator 1 over denominator open parentheses 2 straight k plus 1 close parentheses open parentheses 2 straight k plus 3 close parentheses end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 2 straight k plus 3 end fraction end cell row cell fraction numerator k open parentheses 2 k plus 3 close parentheses over denominator open parentheses 2 straight k plus 1 close parentheses open parentheses 2 straight k plus 3 close parentheses end fraction plus fraction numerator 1 over denominator open parentheses 2 straight k plus 1 close parentheses open parentheses 2 straight k plus 3 close parentheses end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 2 straight k plus 3 end fraction end cell row cell fraction numerator 2 k squared plus 3 k over denominator open parentheses 2 straight k plus 1 close parentheses open parentheses 2 straight k plus 3 close parentheses end fraction plus fraction numerator 1 over denominator open parentheses 2 straight k plus 1 close parentheses open parentheses 2 straight k plus 3 close parentheses end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 2 straight k plus 3 end fraction end cell row cell fraction numerator 2 k squared plus 3 k plus 1 over denominator open parentheses 2 straight k plus 1 close parentheses open parentheses 2 straight k plus 3 close parentheses end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 2 straight k plus 3 end fraction end cell row cell fraction numerator up diagonal strike left parenthesis 2 k plus 1 right parenthesis end strike left parenthesis k plus 1 right parenthesis over denominator up diagonal strike open parentheses 2 straight k plus 1 close parentheses end strike open parentheses 2 straight k plus 3 close parentheses end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 2 straight k plus 3 end fraction end cell row cell fraction numerator straight k plus 1 over denominator 2 straight k plus 3 end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 2 straight k plus 3 end fraction rightwards arrow terbukti end cell end table

Jadi terbukti straight P subscript straight n identical to fraction numerator 1 over denominator 1 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 5 end fraction plus... plus fraction numerator 1 over denominator open parentheses 2 straight n minus 1 close parentheses open parentheses 2 straight n plus 1 close parentheses end fraction equals fraction numerator straight n over denominator 2 straight n plus 1 end fraction karena hasil ruas kiri dan kanan sama

1

Roboguru

Dengan menggunakan induksi matematika, akan dibuktikan bahwa   untuk setiap bilangan asli .  Dengan demikian, dapat disimpulkan bahwa ....

Pembahasan Soal:

Perhatikan pernyataan berikut!

P subscript n colon 1 plus 2 plus midline horizontal ellipsis plus n equals open parentheses n plus 1 half close parentheses squared over 2 

untuk setiap bilangan asli n. Artinya, n greater or equal than 1.

Langkah 1 : Buktikan bold italic P subscript bold 1 benar 

table attributes columnspacing 0 0 end attributes row cell P subscript 1 colon end cell cell 1 equals end cell cell open parentheses 1 plus begin display style 1 half end style close parentheses squared over 2 end cell row blank cell 1 equals end cell cell open parentheses begin display style 3 over 2 end style close parentheses squared over 2 end cell row blank cell 1 equals end cell cell fraction numerator begin display style 9 over 4 end style over denominator 2 end fraction end cell row blank cell 1 equals end cell cell 9 over 8 end cell end table

Perhatikan bahwa 1 not equal to 9 over 8. Dengan demikian P subscript 1 salah.

 

Langkah 2 : Buktikan untuk sembarang bilangan asli bold italic k, jika bold italic P subscript bold k bernilai benar mengakibatkan bold italic P subscript bold k bold plus bold 1 end subscript bernilai benar

Asumsikan P subscript k colon space 1 plus 2 plus midline horizontal ellipsis plus k equals open parentheses k plus 1 half close parentheses squared over 2 bernilai benar.

Akan dibuktikan bahwa P subscript k plus 1 end subscript colon space 1 plus 2 plus... plus open parentheses k plus 1 close parentheses equals open parentheses left parenthesis k plus 1 right parenthesis plus begin display style 1 half end style close parentheses squared over 2 bernilai benar.

Perhatikan perhitungan berikut!

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 plus 2 plus midline horizontal ellipsis plus k plus left parenthesis k plus 1 right parenthesis end cell equals cell 1 plus 2 plus midline horizontal ellipsis plus k plus open parentheses k plus 1 close parentheses end cell row blank equals cell blank over 2 plus left parenthesis k plus 1 right parenthesis end cell row blank equals cell fraction numerator open parentheses k squared plus k plus 1 fourth close parentheses over denominator 2 end fraction plus fraction numerator 2 open parentheses k plus 1 close parentheses over denominator 2 end fraction end cell row blank equals cell fraction numerator open parentheses k squared plus k plus 1 fourth close parentheses over denominator 2 end fraction plus fraction numerator 2 k plus 2 over denominator 2 end fraction end cell row blank equals cell fraction numerator k squared plus 3 k plus begin display style 9 over 4 end style over denominator 2 end fraction end cell row blank equals cell open parentheses k plus begin display style 3 over 2 end style close parentheses squared over 2 end cell row blank equals cell open parentheses left parenthesis k plus 1 right parenthesis plus begin display style 1 half end style close parentheses squared over 2 end cell end table  

Didapat bahwa P subscript k plus 1 end subscript benar.

Oleh karena itu, didapat hasil sebagai berikut.

  1. P subscript 1 tidak bernilai benar.
  2. Untuk sembarang k bilangan asli, jika P subscript k bernilai benar mengakibatkan undefined bernilai benar.

Dengan demikian, pernyataan tidak terbukti karena tahap pertama tidak bisa dibuktikan, meskipun tahap kedua bisa dibuktikan.undefined 

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Gunakan prinsip induksi matematika untuk membuktikan setiap pernyataan berikut  a.

Pembahasan Soal:

Dengan menggunakan induksi matematika dimana

Untuk n = 1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to straight n of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2 straight n plus 3 over denominator 2 to the power of straight n minus 1 end exponent end fraction end cell row cell sum from straight i equals 1 to 1 of fraction numerator 2.1 minus 1 over denominator 2 to the power of 1 minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2.1 plus 3 over denominator 2 to the power of 1 minus 1 end exponent end fraction end cell row cell fraction numerator 2 minus 1 over denominator 2 to the power of 0 end fraction end cell equals cell 6 minus fraction numerator 2 plus 3 over denominator 2 to the power of 0 end fraction end cell row cell 1 over 1 end cell equals cell 6 minus 5 over 1 end cell row 1 equals cell 1 rightwards arrow Terbukti end cell end table

Untuk n = k maka di asumsikan terbukti

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to straight n of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2 straight n plus 3 over denominator 2 to the power of straight n minus 1 end exponent end fraction end cell row cell sum from straight i equals 1 to straight k of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2. straight k plus 3 over denominator 2 to the power of straight k minus 1 end exponent end fraction end cell row cell fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2 straight k plus 3 over denominator 2 to the power of straight k minus 1 end exponent end fraction rightwards arrow Terbukti end cell end table

Untuk n = k+1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to straight n of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2 straight n plus 3 over denominator 2 to the power of straight n minus 1 end exponent end fraction end cell row cell sum from straight i equals 1 to straight k plus 1 of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction end cell equals cell sum from straight i equals 1 to straight k of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction plus sum from straight i equals straight k plus 1 to straight k plus 1 of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction end cell row cell 6 minus fraction numerator 2 left parenthesis straight k plus 1 right parenthesis plus 3 over denominator 2 to the power of straight k plus 1 minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2 straight k plus 3 over denominator 2 to the power of straight k minus 1 end exponent end fraction plus fraction numerator 2 left parenthesis straight k plus 1 right parenthesis minus 1 over denominator 2 to the power of straight k plus 1 minus 1 end exponent end fraction end cell row cell 6 minus fraction numerator 2 straight k plus 2 plus 3 over denominator 2 to the power of straight k end fraction end cell equals cell 6 minus fraction numerator 2 straight k plus 3 over denominator 2 to the power of straight k minus 1 end exponent end fraction plus fraction numerator 2 straight k plus 2 minus 1 over denominator 2 to the power of straight k end fraction end cell row cell 6 minus fraction numerator 2 straight k plus 5 over denominator 2 to the power of straight k end fraction end cell equals cell 6 minus fraction numerator 2 straight k plus 3 over denominator 2 to the power of straight k minus 1 end exponent end fraction plus fraction numerator 2 straight k plus 1 over denominator 2 to the power of straight k end fraction end cell row cell 6 minus fraction numerator 2 straight k plus 5 over denominator 2 to the power of straight k end fraction end cell equals cell 6 minus fraction numerator 1 plus 3 plus 2 straight k plus 1 over denominator 2 to the power of straight k end fraction end cell row cell 6 minus fraction numerator 2 straight k plus 5 over denominator 2 to the power of straight k end fraction end cell equals cell 6 minus fraction numerator 2 straight k plus 5 over denominator 2 to the power of straight k end fraction rightwards arrow Terbukti end cell end table

jadi terbukti bahwa straight P subscript straight n identical to sum from straight i equals 1 to straight n of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction equals 6 minus fraction numerator 2 straight n plus 3 over denominator 2 to the power of straight n minus 1 end exponent end fraction karena hasil sisi kanan dan kiri sama

0

Roboguru

Dengan menggunakan prinsip induksi matematika, buktikanlah: b.

Pembahasan Soal:

Untuk n = 1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to 1 of left parenthesis 2.1 minus 1 right parenthesis left parenthesis 2.1 plus 1 right parenthesis end cell equals cell fraction numerator 4.1 cubed plus 6.1 squared minus 1 over denominator 3 end fraction end cell row cell 1.3 end cell equals cell fraction numerator 4.1 plus 6.1 minus 1 over denominator 3 end fraction end cell row 3 equals cell 3 rightwards arrow terbukti end cell end table

Asumsikan benar untuk n = k Sehingga diperoleh:

i=1k(2i1)(2i+1)=34k3+6k2k

Untuk n = k+1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to straight k plus 1 of left parenthesis 2 straight i minus 1 right parenthesis left parenthesis 2 straight i plus 1 right parenthesis end cell equals cell fraction numerator 4 left parenthesis straight k plus 1 right parenthesis cubed plus 6 left parenthesis straight k plus 1 right parenthesis squared minus left parenthesis straight k plus 1 right parenthesis over denominator 3 end fraction end cell row blank equals cell space fraction numerator left parenthesis 4 straight k cubed plus 12 straight k squared plus 12 straight k plus 4 right parenthesis plus left parenthesis 6 straight k squared plus 12 straight k plus 6 right parenthesis minus left parenthesis straight k plus 1 right parenthesis over denominator 3 end fraction end cell row blank equals cell fraction numerator 4 straight k cubed plus 18 straight k squared plus 23 straight k plus 9 over denominator 3 end fraction end cell end table

Akan dibuktikan

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to straight k plus 1 of left parenthesis 2 straight i minus 1 right parenthesis left parenthesis 2 straight i plus 1 right parenthesis end cell equals cell sum from straight i equals 1 to straight k of left parenthesis 2 straight i minus 1 right parenthesis left parenthesis 2 straight i plus 1 right parenthesis plus sum from straight i equals straight k plus 1 to straight k plus 1 of left parenthesis 2 straight i minus 1 right parenthesis left parenthesis 2 straight i plus 1 right parenthesis end cell row blank equals cell fraction numerator 4 straight k cubed plus 6 straight k squared minus straight k over denominator 3 end fraction plus open parentheses 2 open parentheses straight k plus 1 close parentheses minus 1 close parentheses open parentheses 2 open parentheses straight k plus 1 close parentheses plus 1 close parentheses end cell row blank equals cell fraction numerator 4 straight k cubed plus 6 straight k squared minus straight k over denominator 3 end fraction plus open parentheses 2 straight k plus 2 minus 1 close parentheses open parentheses 2 straight k plus 2 plus 1 close parentheses end cell row blank equals cell fraction numerator 4 straight k cubed plus 6 straight k squared minus straight k over denominator 3 end fraction plus open parentheses 4 straight k squared plus 8 straight k plus 3 close parentheses end cell row blank equals cell fraction numerator 4 straight k cubed plus 6 straight k squared minus straight k plus 12 straight k squared plus 24 straight k plus 9 over denominator 3 end fraction end cell row blank equals cell fraction numerator 4 straight k cubed plus 18 straight k squared plus 23 straight k plus 9 over denominator 3 end fraction rightwards arrow terbukti space end cell end table

Jadi straight P subscript straight n identical to sum from straight i equals 1 to straight n of left parenthesis 2 straight i minus 1 right parenthesis left parenthesis 2 straight i plus 1 right parenthesis equals fraction numerator 4 straight n cubed plus 6 straight n squared minus straight n over denominator 3 end fraction  terbukti menggunkaan induksi matematika karena hasilnya sama

 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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