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Tulislah dalam notasi sigma, kemudian tentukan masing-masing formulanya dalam ekspresi . a. 3+7+11+...+(4n−1)

Pertanyaan

Tulislah dalam notasi sigma, kemudian tentukan masing-masing formulanya dalam ekspresi undefined.

a. 3+7+11+...+(4n1) 

Pembahasan Soal:

Deret pada soal merupakan deret artimetika dengan b=4.

Dari deret aritmatika di atas:

  • Pola ke-n : 4n1  
  • Batas bawah: 

4n14n4nn====33+141 

  • Batas atas: n 

Sehingga notasi sigmanya adalah n=1n(4i1).

  • Formulanya dalam ekspresi undefined.

Sn====2n(a+Un)2n(3+4n1)2n(4n+2)2n2+n 

Dengan demikian, deret dari 3+7+11+...+(4n1) dapat ditulis dalam notasi sigma n=1n(4i1) dan formulanya dalam ekspresi undefined adalah Sn=2n2+n atau dapat ditulis dalam bentuk n=1n(4i1)=2n2+n.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

Buktikan dengan induksi matematika. berlaku untuk semua n bilangan asli.

Pembahasan Soal:

Misalkan begin mathsize 12px style P subscript n identical to 1 half plus 2 over 2 squared plus 3 over 2 cubed plus horizontal ellipsis plus n over 2 to the power of n equals 2 minus fraction numerator n plus 2 over denominator 2 to the power of n end fraction end style.

Langkah-langkah induksi matematika, yaitu sebagai berikut.

Dibuktikan P subscript n benar untuk n equals 1

table attributes columnalign right center left columnspacing 0px end attributes row cell n over 2 to the power of n end cell equals cell 2 minus fraction numerator n plus 2 over denominator 2 to the power of n end fraction end cell row cell 1 over 2 to the power of 1 end cell equals cell 2 minus fraction numerator 1 plus 2 over denominator 2 to the power of 1 end fraction end cell row cell 1 half end cell equals cell 2 minus 3 over 2 end cell row cell 1 half end cell equals cell 1 half end cell end table

Jadi,  P subscript n benar untuk n equals 1

P subscript n diamsusikan benar untuk n equals k, sehingga 

begin mathsize 12px style 1 half plus 2 over 2 squared plus 3 over 2 cubed plus horizontal ellipsis plus k over 2 to the power of k equals 2 minus fraction numerator k plus 2 over denominator 2 to the power of k end fraction end style

P subscript n benar untuk n equals k plus 1

begin mathsize 12px style rightwards arrow Ruas space kiri equals 1 half plus 2 over 2 squared plus 3 over 2 cubed plus horizontal ellipsis plus k over 2 to the power of k plus fraction numerator k plus 1 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 minus fraction numerator k plus 2 over denominator 2 to the power of k end fraction plus fraction numerator k plus 1 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 plus fraction numerator negative 2 left parenthesis k plus 2 right parenthesis plus k plus 1 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 plus fraction numerator negative 2 k minus 4 plus k plus 1 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 plus fraction numerator negative k minus 3 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 plus fraction numerator negative left parenthesis k plus 3 right parenthesis over denominator k to the power of k plus 1 end exponent end fraction equals 2 minus fraction numerator k plus 3 over denominator k to the power of k plus 1 end exponent end fraction end style

begin mathsize 12px style rightwards arrow Ruas space kanan equals 2 minus fraction numerator left parenthesis k plus 1 right parenthesis plus 2 over denominator 2 to the power of left parenthesis k plus 1 right parenthesis end exponent end fraction equals 2 minus fraction numerator k plus 3 over denominator 2 to the power of k plus 1 end exponent end fraction end style

Karena ruas kiri = ruas kanan, maka P subscript nbenar untuk n equals k plus 1 

Dengan demikian,

 begin mathsize 12px style 1 half plus 2 over 2 squared plus 3 over 2 cubed plus horizontal ellipsis plus n over 2 to the power of n equals 2 minus fraction numerator n plus 2 over denominator 2 to the power of n end fraction end style

 terbukti benar untuk n bilangan asli.

1

Roboguru

Buktikan dengan induksi matematika.

Pembahasan Soal:

Prinsip Induksi Matematika:

Misalkan P open parentheses n close parentheses merupakan suatu pernyataan untuk setiap bilangan asli n. Pernyataan P open parentheses n close parentheses benar jika memenuhi langkah berikut.

1. Langkah awal: Dibuktikan P open parentheses 1 close parentheses benar.

2. Langkah induksi: Jika diasumsikan P open parentheses k close parentheses benar, maka harus dibuktikan bahwa P open parentheses k plus 1 close parentheses juga benar, untuk setiap k bilangan asli.

Jika langkah 1 dan 2 sudah diuji kebenarannya, maka ditarik kesimpulan bahwa P open parentheses n close parentheses benar untuk setiap bilangan asli n.

Akan dibuktikan dengan induksi matematika bahwa

sum from k equals 1 to n of k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses equals fraction numerator n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses open parentheses n plus 3 close parentheses over denominator 4 end fraction

Langkah awal:

Akan dibuktikan P open parentheses 1 close parentheses benar.

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 open parentheses 1 plus 1 close parentheses open parentheses 1 plus 2 close parentheses end cell equals cell fraction numerator 1 open parentheses 1 plus 1 close parentheses open parentheses 1 plus 2 close parentheses open parentheses 1 plus 3 close parentheses over denominator 4 end fraction end cell row cell 1 times 2 times 3 end cell equals cell fraction numerator 1 times 2 times 3 times 4 over denominator 4 end fraction end cell row 6 equals 6 end table

Jadi, terbukti P open parentheses 1 close parentheses benar.

Langkah induksi:

P open parentheses n close parentheses diasumsikan benar untuk n equals m sehingga 

sum from k equals 1 to m of k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses equals fraction numerator m open parentheses m plus 1 close parentheses open parentheses m plus 2 close parentheses open parentheses m plus 3 close parentheses over denominator 4 end fraction

Akan ditunjukkan bahwa untuk n equals m plus 1 juga benar, sedemikian sehingga 

sum from k equals 1 to m plus 1 of k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses equals fraction numerator open parentheses m plus 1 close parentheses open parentheses m plus 2 close parentheses open parentheses m plus 3 close parentheses open parentheses m plus 4 close parentheses over denominator 4 end fraction

Bukti:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from k equals 1 to m plus 1 of k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end cell row blank equals cell sum from k equals 1 to m of k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses plus sum from k equals m plus 1 to m plus 1 of k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end cell row blank equals cell fraction numerator m open parentheses m plus 1 close parentheses open parentheses m plus 2 close parentheses open parentheses m plus 3 close parentheses over denominator 4 end fraction plus open parentheses m plus 1 close parentheses open parentheses m plus 2 close parentheses open parentheses m plus 3 close parentheses end cell row blank equals cell fraction numerator m open parentheses m plus 1 close parentheses open parentheses m plus 2 close parentheses open parentheses m plus 3 close parentheses over denominator 4 end fraction plus fraction numerator 4 open parentheses m plus 1 close parentheses open parentheses m plus 2 close parentheses open parentheses m plus 3 close parentheses over denominator 4 end fraction end cell row blank equals cell fraction numerator open parentheses m plus 1 close parentheses open parentheses m plus 2 close parentheses open parentheses m plus 3 close parentheses open parentheses m plus 4 close parentheses over denominator 4 end fraction end cell end table

Jadi, terbukti bahwa P open parentheses n close parentheses benar untuk n equals m plus 1.

Pernyataan P open parentheses n close parentheses memenuhi kedua prinsip induksi matematika.

Dengan demikian, berdasarkan prinsip induksi matematika, P open parentheses n close parentheses benar untuk setiap n bilangan asli.

2

Roboguru

Buktikan dengan induksi matematika.

Pembahasan Soal:

Misalkan P open parentheses n close parentheses colon left parenthesis 1 cross times 2 right parenthesis plus left parenthesis 2 cross times 3 right parenthesis plus left parenthesis 3 cross times 4 right parenthesis plus horizontal ellipsis plus n left parenthesis n plus 1 right parenthesis equals fraction numerator n left parenthesis n plus 1 right parenthesis left parenthesis n plus 2 right parenthesis over denominator 3 end fraction .

Langkah-langkah induksi matematika, yaitu sebagai berikut.

1. Langkah dasar/awal : Tunjukkan P open parentheses 1 close parentheses benar.

Untuk n equals 1 rightwards arrow P open parentheses 1 close parentheses colon 

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 cross times 2 end cell equals cell fraction numerator n left parenthesis n plus 1 right parenthesis left parenthesis n plus 2 right parenthesis over denominator 3 end fraction end cell row 2 equals cell fraction numerator 1 left parenthesis 1 plus 1 right parenthesis left parenthesis 1 plus 2 right parenthesis over denominator 3 end fraction end cell row 2 equals cell fraction numerator 1 times 2 times 3 over denominator 3 end fraction end cell row 2 equals cell 6 over 3 end cell row 2 equals cell 2 rightwards arrow open parentheses benar close parentheses end cell end table 

2. Asumsikan P open parentheses k close parentheses benar untuk sembarang k bilangan asli, kemudian tunjukkan P open parentheses k plus 1 close parentheses juga benar berdasarkan asumsi tersebut.

# Asumsikan bahwa P open parentheses k close parentheses benar  

left parenthesis 1 cross times 2 right parenthesis plus left parenthesis 2 cross times 3 right parenthesis plus left parenthesis 3 cross times 4 right parenthesis plus horizontal ellipsis plus k left parenthesis k plus 1 right parenthesis equals fraction numerator k left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis over denominator 3 end fraction 

# Akan menunjukkan P open parentheses k plus 1 close parentheses benar 

left parenthesis 1 cross times 2 right parenthesis plus horizontal ellipsis plus k left parenthesis k plus 1 right parenthesis plus left parenthesis k plus 1 right parenthesis left parenthesis left parenthesis k plus 1 right parenthesis plus 1 right parenthesis equals fraction numerator open parentheses k plus 1 close parentheses open parentheses k plus 1 plus 1 close parentheses open parentheses k plus 1 plus 2 close parentheses over denominator 3 end fraction space space space space space space space space space space space space space space space space space space fraction numerator straight k left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis over denominator 3 end fraction plus left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis equals fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis left parenthesis k plus 3 right parenthesis over denominator 3 end fraction space space space space space space space space space space space space space space space fraction numerator straight k left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis over denominator 3 end fraction plus fraction numerator 3 left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis over denominator 3 end fraction equals fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis left parenthesis k plus 3 right parenthesis over denominator 3 end fraction 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 fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis left parenthesis k plus 3 right parenthesis over denominator 3 end fraction equals fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis left parenthesis k plus 3 right parenthesis over denominator 3 end fraction rightwards arrow open parentheses benar close parentheses 

 

3. Kesimpulan : Terbukti bahwa left parenthesis 1 cross times 2 right parenthesis plus left parenthesis 2 cross times 3 right parenthesis plus left parenthesis 3 cross times 4 right parenthesis plus horizontal ellipsis plus n left parenthesis n plus 1 right parenthesis equals fraction numerator n left parenthesis n plus 1 right parenthesis left parenthesis n plus 2 right parenthesis over denominator 3 end fraction benar untuk setiap bilangan asli n.

Dengan demikian, Terbukti bahwa left parenthesis 1 cross times 2 right parenthesis plus left parenthesis 2 cross times 3 right parenthesis plus left parenthesis 3 cross times 4 right parenthesis plus horizontal ellipsis plus n left parenthesis n plus 1 right parenthesis equals fraction numerator n left parenthesis n plus 1 right parenthesis left parenthesis n plus 2 right parenthesis over denominator 3 end fraction benar untuk setiap bilangan asli n.

3

Roboguru

Buktikan dengan induksi matematika.

Pembahasan Soal:

Misalkan P subscript n identical to sum from i equals 1 to n of 3 to the power of i minus 1 end exponent equals 1 half open parentheses 3 to the power of n minus 1 close parentheses.

Langkah-langkah induksi matematika, yaitu sebagai berikut.

Dibuktikan P subscript n benar untuk n equals 1

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 1 to n of 3 to the power of i minus 1 end exponent end cell equals cell 1 half open parentheses 3 to the power of n minus 1 close parentheses end cell row cell 3 to the power of 1 minus 1 end exponent end cell equals cell 1 half open parentheses 3 to the power of 1 minus 1 close parentheses end cell row cell 3 to the power of 0 end cell equals cell 1 half open parentheses 2 close parentheses end cell row 1 equals cell 1 space left parenthesis Benar right parenthesis end cell end table

Jadi,  P subscript n benar untuk n equals 1

P subscript n diamsusikan benar untuk n equals k, sehingga 

sum from i equals 1 to k of 3 to the power of i minus 1 end exponent equals 1 half open parentheses 3 to the power of k minus 1 close parentheses

Akan dibuktikan P subscript n benar untuk n equals k plus 1

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 1 to k plus 1 of 3 to the power of i minus 1 end exponent end cell equals cell 1 half open parentheses 3 to the power of k plus 1 end exponent minus 1 close parentheses end cell row cell sum from i equals 1 to k of 3 to the power of i minus 1 end exponent plus 3 to the power of left parenthesis k plus 1 right parenthesis minus 1 end exponent end cell equals cell 1 half open parentheses 3 cross times 3 to the power of k minus 1 close parentheses end cell row cell 1 half left parenthesis 3 to the power of k minus 1 right parenthesis plus 3 to the power of k end cell equals cell fraction numerator 3 cross times 3 to the power of k minus 1 over denominator 2 end fraction end cell row cell fraction numerator 3 to the power of k minus 1 plus 2 cross times 3 to the power of k over denominator 2 end fraction end cell equals cell fraction numerator 3 cross times 3 to the power of k minus 1 over denominator 2 end fraction end cell row cell fraction numerator 3 cross times 3 to the power of k minus 1 over denominator 2 end fraction end cell equals cell fraction numerator 3 cross times 3 to the power of k minus 1 over denominator 2 end fraction end cell end table

Dengan demikian,

 P subscript n identical to sum from i equals 1 to n of 3 to the power of i minus 1 end exponent equals 1 half open parentheses 3 to the power of n minus 1 close parentheses

 terbukti benar untuk semua n bilangan asli.

2

Roboguru

Buktikan dengan induksi matematika. berlaku untuk semua n bilangan asli.

Pembahasan Soal:

Misalkan 

begin mathsize 12px style P subscript n identical to left parenthesis 1 cross times 2 right parenthesis plus left parenthesis 2 cross times 3 right parenthesis plus left parenthesis 3 cross times 4 right parenthesis plus left parenthesis 4 cross times 5 right parenthesis plus horizontal ellipsis plus n left parenthesis n plus 1 right parenthesis equals n over 3 left parenthesis n plus 1 right parenthesis left parenthesis n plus 2 right parenthesis end style.

Langkah-langkah induksi matematika, yaitu sebagai berikut.

Dibuktikan P subscript n benar untuk n equals 1

table attributes columnalign right center left columnspacing 0px end attributes row cell n left parenthesis n plus 1 right parenthesis end cell equals cell fraction numerator n left parenthesis n plus 1 right parenthesis left parenthesis n plus 2 right parenthesis over denominator 3 end fraction end cell row cell 1 left parenthesis 1 plus 1 right parenthesis end cell equals cell fraction numerator 1 left parenthesis 1 plus 1 right parenthesis left parenthesis 1 plus 2 right parenthesis over denominator 3 end fraction end cell row 2 equals cell fraction numerator left parenthesis 2 right parenthesis left parenthesis 3 right parenthesis over denominator 3 end fraction end cell row 2 equals cell 6 over 3 end cell row 2 equals 2 end table

Jadi,  P subscript n benar untuk n equals 1

P subscript n diamsusikan benar untuk n equals k, sehingga 

begin mathsize 12px style left parenthesis 1 cross times 2 right parenthesis plus left parenthesis 2 cross times 3 right parenthesis plus left parenthesis 3 cross times 4 right parenthesis plus left parenthesis 4 cross times 5 right parenthesis plus horizontal ellipsis plus k left parenthesis k plus 1 right parenthesis equals k over 3 left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis end style

P subscript n benar untuk n equals k plus 1

begin mathsize 12px style rightwards arrow Ruas space kiri equals left parenthesis 1 cross times 2 right parenthesis plus horizontal ellipsis plus k left parenthesis k plus 1 right parenthesis plus left parenthesis k plus 1 right parenthesis left parenthesis left parenthesis k plus 1 right parenthesis plus 1 right parenthesis equals fraction numerator straight k left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis over denominator 3 end fraction plus left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis equals 1 third open parentheses straight k left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis close parentheses plus 1 third 3 left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis equals 1 third left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis left parenthesis straight k plus 3 right parenthesis end style

begin mathsize 12px style rightwards arrow Ruas space kanan equals fraction numerator k left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis over denominator 3 end fraction equals fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis left parenthesis k plus 1 right parenthesis plus 1 right parenthesis left parenthesis left parenthesis k plus 1 right parenthesis plus 2 right parenthesis over denominator 3 end fraction equals 1 third left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis left parenthesis k plus 3 right parenthesis end style

Karena ruas kiri = ruas kanan, maka P subscript nbenar untuk n equals k plus 1 

Dengan demikian,

 begin mathsize 12px style left parenthesis 1 cross times 2 right parenthesis plus left parenthesis 2 cross times 3 right parenthesis plus left parenthesis 3 cross times 4 right parenthesis plus horizontal ellipsis plus n left parenthesis n plus 1 right parenthesis equals fraction numerator n left parenthesis n plus 1 right parenthesis left parenthesis n plus 2 right parenthesis over denominator 3 end fraction end style

 terbukti benar untuk bilangan asli.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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