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Gunakan prinsip induksi matematika untuk membuktikan setiap notasi sigma berikut. c.

Pertanyaan

Gunakan prinsip induksi matematika untuk membuktikan setiap notasi sigma berikut.

c. sum from straight j equals 1 to straight n of fraction numerator 1 over denominator open parentheses 3 straight j minus 2 close parentheses open parentheses 3 straight j plus 1 close parentheses end fraction equals fraction numerator straight n over denominator 3 straight n plus 1 end fraction 

Pembahasan Soal:

Pembuktian dengan menggunakan induksi matematika dimana untuk n = 1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight j equals 1 to straight n of fraction numerator 1 over denominator open parentheses 3 straight j minus 2 close parentheses open parentheses 3 straight j plus 1 close parentheses end fraction end cell equals cell fraction numerator straight n over denominator 3 straight n plus 1 end fraction end cell row cell fraction numerator 1 over denominator open parentheses 3.1 minus 2 close parentheses open parentheses 3.1 plus 1 close parentheses end fraction end cell equals cell fraction numerator 1 over denominator 3.1 plus 1 end fraction end cell row cell fraction numerator 1 over denominator 1.4 end fraction end cell equals cell 1 fourth end cell row cell 1 fourth end cell equals cell 1 fourth rightwards arrow terbukti end cell end table

Untuk n = k akan diasumsikan terbukti maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight j equals 1 to straight k of fraction numerator 1 over denominator open parentheses 3 straight j minus 2 close parentheses open parentheses 3 straight j plus 1 close parentheses end fraction end cell equals cell fraction numerator straight k over denominator 3 straight k plus 1 end fraction rightwards arrow terbukti end cell end table

Untuk n = k+1 maka dibuktikan

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight j equals 1 to straight k plus 1 of fraction numerator 1 over denominator open parentheses 3 straight j minus 2 close parentheses open parentheses 3 straight j plus 1 close parentheses end fraction end cell equals cell sum from straight j equals 1 to straight k of fraction numerator 1 over denominator open parentheses 3 straight j minus 2 close parentheses open parentheses 3 straight j plus 1 close parentheses end fraction plus sum from straight j equals straight k plus 1 to straight k plus 1 of fraction numerator 1 over denominator open parentheses 3 straight j minus 2 close parentheses open parentheses 3 straight j plus 1 close parentheses end fraction end cell row cell fraction numerator straight k plus 1 over denominator 3 left parenthesis straight k plus 1 right parenthesis plus 1 end fraction end cell equals cell fraction numerator straight k over denominator 3 straight k plus 1 end fraction plus fraction numerator 1 over denominator open parentheses 3 left parenthesis straight k plus 1 right parenthesis minus 2 close parentheses open parentheses 3 open parentheses straight k plus 1 close parentheses plus 1 close parentheses end fraction end cell row cell fraction numerator straight k plus 1 over denominator 3 straight k plus 3 plus 1 end fraction end cell equals cell fraction numerator straight k over denominator 3 straight k plus 1 end fraction plus fraction numerator 1 over denominator open parentheses 3 straight k plus 3 minus 2 close parentheses open parentheses 3 straight k plus 3 plus 1 close parentheses end fraction end cell row cell fraction numerator straight k plus 1 over denominator 3 straight k plus 4 end fraction end cell equals cell fraction numerator straight k open parentheses 3 straight k plus 4 close parentheses plus 1 over denominator open parentheses 3 straight k plus 1 close parentheses open parentheses 3 straight k plus 4 close parentheses end fraction end cell row cell fraction numerator straight k plus 1 over denominator 3 straight k plus 4 end fraction end cell equals cell fraction numerator 3 straight k squared plus 4 straight k plus 1 over denominator open parentheses 3 straight k plus 1 close parentheses open parentheses 3 straight k plus 4 close parentheses end fraction end cell row cell fraction numerator straight k plus 1 over denominator 3 straight k plus 4 end fraction end cell equals cell fraction numerator open parentheses 3 straight k plus 4 close parentheses left parenthesis straight k plus 1 right parenthesis over denominator open parentheses 3 straight k plus 1 close parentheses open parentheses 3 straight k plus 4 close parentheses end fraction end cell row cell fraction numerator straight k plus 1 over denominator 3 straight k plus 4 end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 3 straight k plus 4 end fraction rightwards arrow terbukti end cell end table

Jadi terbukti bahwa sum from straight j equals 1 to straight n of fraction numerator 1 over denominator open parentheses 3 straight j minus 2 close parentheses open parentheses 3 straight j plus 1 close parentheses end fraction equals fraction numerator straight n over denominator 3 straight n plus 1 end fraction karena hasil sisi kiri dan kanan sama

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 18 Juli 2021

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