Roboguru

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

Pertanyaan

Gunakan prinsip induksi matematika untuk membuktikan setiap notasi sigma berikut.

d. sum from straight i equals 1 to straight n of straight i over 2 to the power of straight i equals 2 minus fraction numerator straight n plus 2 over denominator 2 to the power of straight n end fraction 

Pembahasan Soal:

Pembuktian dengan 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 straight i over 2 to the power of straight i end cell equals cell 2 minus fraction numerator straight n plus 2 over denominator 2 to the power of straight n end fraction end cell row cell sum from straight i equals 1 to 1 of 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 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 i equals 1 to straight n of straight i over 2 to the power of straight i end cell equals cell 2 minus fraction numerator straight n plus 2 over denominator 2 to the power of straight n end fraction end cell row cell sum from straight i equals 1 to straight k of 1 over 2 to the power of 1 end cell equals cell 2 minus fraction numerator straight k plus 2 over denominator 2 to the power of straight k end fraction rightwards arrow terbukti end cell end table

Untuk n = k+1

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to straight n of straight i over 2 to the power of straight i end cell equals cell 2 minus fraction numerator straight n plus 2 over denominator 2 to the power of straight n end fraction end cell row cell sum from straight i equals 1 to straight k plus 1 of 1 over 2 to the power of straight i end cell equals cell sum from straight i equals 1 to straight k of 1 over 2 to the power of straight i plus fraction numerator straight k plus 1 over denominator 2 to the power of straight k plus 1 end exponent end fraction end cell row blank equals cell 2 minus fraction numerator straight k plus 2 over denominator 2 to the power of straight k end fraction plus fraction numerator straight k plus 1 over denominator 2 to the power of straight k plus 1 end exponent end fraction end cell row blank equals cell 2 minus open square brackets fraction numerator 2 open parentheses straight k plus 2 close parentheses minus open parentheses straight k plus 1 close parentheses over denominator 2 to the power of straight k plus 1 end exponent end fraction close square brackets end cell row blank equals cell 2 minus open square brackets fraction numerator 2 straight k plus 4 minus straight k minus 1 over denominator 2 to the power of straight k plus 1 end exponent end fraction close square brackets end cell row blank equals cell 2 minus open square brackets fraction numerator straight k plus 3 over denominator 2 to the power of straight k plus 1 end exponent end fraction close square brackets end cell row blank equals cell 2 minus fraction numerator open parentheses straight k plus 1 close parentheses plus 2 over denominator 2 to the power of straight k plus 1 end exponent end fraction rightwards arrow terbu kti end cell end table

Jadi terbukti bahwa sum from straight i equals 1 to straight n of straight i over 2 to the power of straight i equals 2 minus fraction numerator straight n plus 2 over denominator 2 to the power of straight n end fraction karena hasil sisi kanan dan kiri sama

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 18 Juli 2021

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