Roboguru

Buktikan dengan prinsip induksi matematika.  b.

Pertanyaan

Buktikan dengan prinsip induksi matematika. 

b. begin mathsize 14px style sum from i equals 1 to n of 3 over n open square brackets open parentheses i over n close parentheses squared plus 1 close square brackets equals fraction numerator open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 2 n squared end fraction plus 3 end style 

Pembahasan Soal:

Langkah-langkah induksi:

  1. Buktikan untuk bilangan 1, pernyataan tersebut benar.
  2. Nyatakan untuk bilangan asli sembarang, misalnya k, pernyataan tersebut diasumsikan benar.
  3. Buktikan untuk bilangan asli k plus 1 pernyataan tersebut juga benar. 

Maka:

Langkah 1: 

Untuk n equals 1, maka

13[(11)2+1]3×26===2(1)2(1+1)(2(1)+1)+322×3+36   

Pernyataan benar untuk n equals 1.

Langkah 2:

Nyatakan untuk bilangan asli sembarang, misalnya k, pernyataan tersebut diasumsikan benar. sehingga:

i=1kk3[(ki)2+1]=2k2(k+1)(2k+1)+3   

Langkah 3:

Untuk n equals k plus 1, maka

i=1k+1k+13[(k+1i)2+1]=2(k+1)2(k+1)+1)(2(k+1)+1)+3        

Pada tahap ini, Untuk n equals k plus 1 dan n=k memiliki deret yang berbeda, sehingga untuk i=1nn3[(ni)2+1]=2n2(n+1)(2n+1)+3 tidak dapat dibuktikan.

Dengan demikian, i=1nn3[(ni)2+1]=2n2(n+1)(2n+1)+3 merupakan pernyataan yang salah.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 12 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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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

Prove by induction that

Pembahasan Soal:

Untuk n = 1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight k equals 1 to straight n of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent end cell equals cell straight n times 2 to the power of straight n end cell row cell sum from straight k equals 1 to 1 of open parentheses 1 plus 1 close parentheses times 2 to the power of 1 minus 1 end exponent end cell equals cell 1 times 2 to the power of 1 end cell row cell 2.1 end cell equals cell 1.2 end cell row 2 equals cell 2 rightwards arrow Terbukti space end cell end table

Untuk n = k diasumsikan terbukti maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight k equals 1 to straight n of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent end cell equals cell straight n times 2 to the power of straight n end cell row cell sum from straight k equals 1 to straight k of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent end cell equals cell straight k times 2 to the power of straight k end cell row blank equals cell 2 straight k to the power of straight k 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 k equals 1 to straight n of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent end cell equals cell straight n times 2 to the power of straight n end cell row cell sum from straight k equals 1 to straight k of open parentheses straight k plus 1 plus 1 close parentheses times 2 to the power of straight k plus 1 minus 1 end exponent end cell equals cell left parenthesis straight k plus 1 right parenthesis times 2 to the power of left parenthesis straight k plus 1 right parenthesis end exponent end cell row cell open parentheses straight k plus 2 close parentheses.2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 2 to the power of straight k plus 1 end exponent end cell row cell 2 straight k to the power of straight k plus 4 to the power of straight k plus 2 straight k to the power of straight k end cell equals cell 2 straight k to the power of straight k.2 straight k to the power of 1 plus 2 to the power of straight k.2 to the power of 1 end cell row cell 4 straight k to the power of straight k plus 4 to the power of straight k end cell equals cell 4 straight k to the power of straight k plus 4 to the power of straight k rightwards arrow Terbukti end cell end table

Jadi terbukti untuk sum from straight k equals 1 to straight n of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent equals straight n times 2 to the power of straight n karena hasil dari sisi kiri dan kanan sama

1

Roboguru

Buktikan dengan induksi matematika bahwa   berlaku untuk semua  bilangan asli

Pembahasan Soal:

Misalkan undefined adalah 3 plus 6 plus 10 plus 15 plus... plus 1 half open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses equals 1 over 6 n open parentheses n squared plus 6 n plus 11 close parentheses berlaku untuk semua bilangan asli.

Langkah 1

Akan dibuktikan undefined untuk  begin mathsize 14px style n equals 1 end style.

table attributes columnalign right center left columnspacing 0px end attributes row cell P open parentheses n close parentheses colon 3 end cell equals cell 1 over 6 n open parentheses n squared plus 6 n plus 11 close parentheses end cell row blank equals cell 1 over 6 cross times 1 open parentheses 1 squared plus 6 cross times 1 plus 11 close parentheses end cell row blank equals cell 1 over 6 open parentheses 1 plus 6 plus 11 close parentheses end cell row blank equals cell 1 over 6 open parentheses 18 close parentheses end cell row blank equals 3 end table 

Karena ruas kiri sama dengan ruas kana, maka undefined benar untuk undefined.

Langkah 2

Misalkan undefined benar untuk begin mathsize 14px style n equals k end style, maka akan dibuktikan undefined benar untuk begin mathsize 14px style n equals k plus 1 end style.

3 plus 6 plus 10 plus 15 plus... plus 1 half open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses equals 1 over 6 k open parentheses k squared plus 6 k plus 11 close parentheses

Bukti ruas kiri :

3 plus 6 plus 10 plus 15 plus... plus 1 half open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses plus 1 half open parentheses open parentheses k plus 1 close parentheses plus 1 close parentheses open parentheses open parentheses k plus 1 close parentheses plus 2 close parentheses equals 1 over 6 k open parentheses k squared plus 6 k plus 11 close parentheses plus 1 half open parentheses open parentheses k plus 1 close parentheses plus 1 close parentheses open parentheses open parentheses k plus 1 close parentheses plus 2 close parentheses equals 1 over 6 k open parentheses k squared plus 6 k plus 11 close parentheses plus 1 half open parentheses k plus 2 close parentheses open parentheses k plus 3 close parentheses equals 1 over 6 k open parentheses k squared plus 6 k plus 11 close parentheses plus 3 over 6 open parentheses k squared plus 5 k plus 6 close parentheses equals 1 over 6 k cubed plus k squared plus fraction numerator 11 k over denominator 6 end fraction plus 3 over 6 k squared plus 15 over 6 k plus 3 equals 1 over 6 k cubed plus 9 over 6 k squared plus 26 over 6 k plus 3 equals 1 over 6 open parentheses k cubed plus 9 k squared plus 26 k plus 18 close parentheses equals 1 over 6 open parentheses k plus 1 close parentheses open parentheses k squared plus 8 k plus 18 close parentheses equals 1 over 6 open parentheses k plus 1 close parentheses open square brackets open parentheses k squared plus 2 k plus 1 close parentheses plus 6 k plus 17 close square brackets equals 1 over 6 open parentheses k plus 1 close parentheses open square brackets open parentheses k plus 1 close parentheses squared plus 6 k plus 6 plus 11 close square brackets equals 1 over 6 open parentheses k plus 1 close parentheses open square brackets open parentheses k plus 1 close parentheses squared plus 6 open parentheses k plus 1 close parentheses plus 11 close square brackets 

Bukti ruas kanan

begin mathsize 14px style 1 over 6 n open parentheses n squared plus 6 n plus 11 close parentheses space text untuk end text space n equals k plus 1 equals 1 over 6 open parentheses k plus 1 close parentheses open square brackets open parentheses k plus 1 close parentheses squared plus 6 open parentheses k plus 1 close parentheses plus 11 close square brackets end style 

Karena ruas kiri dan ruas kana sama maka undefined untuk undefined terbukti bernilai benar

Jadi langkah 1 dan langkah 2 bernilai benar, maka undefined berlaku untuk setiap bilangan asli

1

Roboguru

Buktikan dengan induksi matematika sederhana bahwa untuk setiap  bilangan asli berlaku:

Pembahasan Soal:

Misalkan undefined adalah rumus begin mathsize 14px style fraction numerator 1 over denominator 1 cross times 2 end fraction plus fraction numerator 1 over denominator 2 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 4 end fraction plus... plus fraction numerator 1 over denominator n left parenthesis n plus 1 right parenthesis end fraction equals fraction numerator n over denominator n plus 1 end fraction end style .

Langkah 1:

Akan dibuktikan undefined benar untuk undefined. Dengan mensubstitusikan undefined ke kedua ruas diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator 1 cross times 2 end fraction end cell equals cell fraction numerator 1 over denominator 1 plus 1 end fraction end cell row cell 1 half end cell equals cell 1 half end cell row cell ruas space kiri end cell equals cell ruas space kanan end cell end table end style    

Oleh karena undefined, maka undefined benar untuk undefined.

Langkah 2:

Andaikan undefined benar untuk undefined, yaitu begin mathsize 14px style sum from m equals 1 to k of fraction numerator 1 over denominator m left parenthesis m plus 1 right parenthesis end fraction equals fraction numerator k over denominator k plus 1 end fraction end style bernilai benar, maka akan dibuktikan undefined benar untuk undefined, yaitu: 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from m equals 1 to k plus 1 of fraction numerator 1 over denominator m left parenthesis m plus 1 right parenthesis end fraction end cell equals cell fraction numerator k plus over denominator left parenthesis k plus 1 right parenthesis plus 1 end fraction end cell row blank equals cell fraction numerator k plus 1 over denominator k plus 2 end fraction end cell end table end style 

Bukti:

Untuk undefined, ruas kiri menjadi:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from m equals 1 to k plus 1 of fraction numerator 1 over denominator m left parenthesis m plus 1 right parenthesis end fraction end cell equals cell sum from m equals 1 to k of fraction numerator 1 over denominator m left parenthesis m plus 1 right parenthesis end fraction plus sum from m equals k plus 1 to k plus 1 of fraction numerator 1 over denominator m left parenthesis m plus 1 right parenthesis end fraction end cell row blank equals cell fraction numerator k over denominator k plus 1 end fraction plus fraction numerator 1 over denominator left parenthesis k plus 1 right parenthesis left parenthesis k plus 1 plus 1 right parenthesis end fraction end cell row blank equals cell fraction numerator k over denominator k plus 1 end fraction plus fraction numerator 1 over denominator left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis end fraction end cell row blank equals cell fraction numerator k left parenthesis k plus 2 right parenthesis plus k plus 1 over denominator left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis end fraction end cell row blank equals cell fraction numerator k squared plus 2 k plus 1 over denominator left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis end fraction end cell row blank equals cell fraction numerator k plus 1 over denominator k plus 2 end fraction end cell row blank equals cell ruas space kanan end cell end table end style    

Oleh karena undefined, maka terbukti bahwa begin mathsize 14px style sum from m equals 1 to k plus 1 of fraction numerator 1 over denominator m left parenthesis m plus 1 right parenthesis end fraction equals fraction numerator k plus 1 over denominator k plus 2 end fraction end style.

Oleh karena Langkah 1 dan Langkah 2 keduanya bernilai benar, maka terbukti bahwa begin mathsize 14px style fraction numerator 1 over denominator 1 cross times 2 end fraction plus fraction numerator 1 over denominator 2 cross times 3 end fraction plus fraction numerator 1 over denominator 3 cross times 4 end fraction plus... plus fraction numerator 1 over denominator n left parenthesis n plus 1 right parenthesis end fraction equals fraction numerator n over denominator n plus 1 end fraction end style untuk setiap undefined bilangan asli.

0

Roboguru

Buktikanlah. a.

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 straight n of straight a to the power of straight n minus 1 end exponent times straight b to the power of straight n minus 1 end exponent end cell equals cell fraction numerator straight a to the power of straight n minus straight b to the power of straight n over denominator straight a minus straight b end fraction end cell row cell sum from straight i equals 1 to 1 of straight a to the power of 1 minus 1 end exponent times straight b to the power of 1 minus 1 end exponent end cell equals cell fraction numerator straight a to the power of 1 minus straight b to the power of 1 over denominator straight a minus straight b end fraction end cell row cell 1.1 end cell equals cell fraction numerator straight a minus straight b over denominator straight a minus straight b end fraction end cell row 1 equals cell 1 rightwards arrow benar space end cell row blank blank blank end table

Untuk n = k diasumsikan benar maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to straight n of straight a to the power of straight n minus 1 end exponent times straight b to the power of straight n minus 1 end exponent end cell equals cell fraction numerator straight a to the power of straight n minus straight b to the power of straight n over denominator straight a minus straight b end fraction end cell row cell sum from straight i equals 1 to 1 of straight a to the power of straight n minus 1 end exponent times straight b to the power of straight n minus 1 end exponent end cell equals cell fraction numerator straight a to the power of straight k minus straight b to the power of straight k over denominator straight a minus straight b end fraction rightwards arrow benar end cell row blank blank blank end table

Akan dibuktikan 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 straight a to the power of straight n minus 1 end exponent times straight b to the power of straight n minus 1 end exponent end cell equals cell fraction numerator straight a to the power of straight n minus straight b to the power of straight n over denominator straight a minus straight b end fraction end cell row cell sum from straight i equals 1 to k plus 1 of straight a to the power of k plus 1 minus 1 end exponent times straight b to the power of k plus 1 minus 1 end exponent end cell equals cell sum from straight i equals 1 to k of straight a to the power of k minus 1 end exponent times straight b to the power of k minus 1 end exponent plus sum from straight i equals straight k plus 1 to straight k plus 1 of straight a to the power of straight k plus 1 minus 1 end exponent times straight b to the power of straight k plus 1 minus 1 end exponent end cell row cell fraction numerator straight a to the power of straight k plus 1 end exponent minus straight b to the power of straight k plus 1 end exponent over denominator straight a minus straight b end fraction end cell equals cell fraction numerator straight a to the power of straight k minus straight b to the power of straight k over denominator straight a minus straight b end fraction plus straight a to the power of straight k. straight b to the power of straight k end cell row cell fraction numerator straight a to the power of straight k plus 1 end exponent minus straight b to the power of straight k plus 1 end exponent over denominator straight a minus straight b end fraction end cell equals cell fraction numerator straight a to the power of straight k minus straight b to the power of straight k over denominator straight a minus straight b end fraction plus fraction numerator open parentheses straight a minus straight b close parentheses straight a to the power of straight k. straight b to the power of straight k over denominator straight a minus straight b end fraction end cell row cell fraction numerator straight a to the power of straight k plus 1 end exponent minus straight b to the power of straight k plus 1 end exponent over denominator straight a minus straight b end fraction end cell equals cell fraction numerator straight a to the power of straight k minus straight b to the power of straight k over denominator straight a minus straight b end fraction plus fraction numerator open parentheses straight a minus straight b close parentheses straight a to the power of straight k. straight b to the power of straight k over denominator straight a minus straight b end fraction end cell row cell fraction numerator straight a to the power of straight k plus 1 end exponent minus straight b to the power of straight k plus 1 end exponent over denominator straight a minus straight b end fraction end cell equals cell fraction numerator straight a to the power of straight k plus 1 end exponent minus straight b to the power of straight k plus 1 end exponent over denominator straight a minus straight b end fraction rightwards arrow benar end cell row blank blank blank end table

Jadi terbukti bahwa sum from straight i equals 1 to straight n of straight a to the power of straight n minus 1 end exponent times straight b to the power of straight n minus 1 end exponent equals fraction numerator straight a to the power of straight n minus straight b to the power of straight n over denominator straight a minus straight b end fraction karena hasil sisi kira dan kanan sama

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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