Roboguru

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

Pertanyaan

Gunakan prinsip induksi matematika untuk membuktikan setiap notasi sigma berikut.

b. sum from straight p equals 1 to straight n of fraction numerator 1 over denominator 4 straight p squared minus 1 end fraction equals fraction numerator 1 over denominator 2 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 p equals 1 to straight n of fraction numerator 1 over denominator 4 straight p squared minus 1 end fraction end cell equals cell fraction numerator 1 over denominator 2 straight n plus 1 end fraction end cell row cell sum from straight p equals 1 to 1 of fraction numerator 1 over denominator 4.1 squared minus 1 end fraction end cell equals cell fraction numerator 1 over denominator 2.1 plus 1 end fraction end cell row cell 1 third end cell equals cell 1 third 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 sum from straight p equals 1 to straight k of fraction numerator 1 over denominator 4 straight p squared minus 1 end fraction end cell equals cell fraction numerator 1 over denominator 2 straight k plus 1 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 p equals 1 to straight n of fraction numerator 1 over denominator 4 straight p squared minus 1 end fraction end cell equals cell fraction numerator 1 over denominator 2 straight n plus 1 end fraction end cell row cell sum from straight p equals 1 to straight k plus 1 of fraction numerator 1 over denominator 4 straight p squared minus 1 end fraction end cell equals cell sum from straight p equals 1 to straight k of fraction numerator 1 over denominator 4 straight p squared minus 1 end fraction plus fraction numerator 1 over denominator 4 open parentheses straight k plus 1 close parentheses squared minus 1 end fraction end cell row blank equals cell fraction numerator 1 over denominator 2 straight k plus 1 end fraction plus fraction numerator 1 over denominator 4 open parentheses straight k squared plus 2 straight k plus 1 close parentheses minus 1 end fraction end cell row blank equals cell fraction numerator 1 over denominator 2 straight k plus 1 end fraction plus fraction numerator 1 over denominator 8 straight k squared plus 8 straight k plus 4 minus 1 end fraction end cell row blank equals cell fraction numerator 1 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 row blank equals cell fraction numerator open parentheses 2 straight k plus 3 close parentheses 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 row blank equals cell fraction numerator open parentheses 2 straight k plus 4 close parentheses over denominator open parentheses 2 straight k plus 1 close parentheses open parentheses 2 straight k plus 3 close parentheses end fraction rightwards arrow Terbukti end cell end table

jadi terbukti bahwa sum from straight p equals 1 to straight n of fraction numerator 1 over denominator 4 straight p squared minus 1 end fraction equals fraction numerator 1 over denominator 2 straight n plus 1 end fraction karena hasil sisi kanan dan kiri sama

 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 18 Juli 2021

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