Pertanyaan

Buktikanlah. d. k = 1 ∑ n 2 k − 1 2 k − 1 = 6 − 2 n − 1 2 n + 3

Buktikanlah.

d. $k = 1 \sum n \frac{2 k - 1}{2 ^{k - 1}} = 6 - \frac{2 n + 3}{2 ^{n - 1}}$

A. Acfreelance

Master Teacher

Mahasiswa/Alumni UIN Walisongo Semarang

Jawaban terverifikasi

Pembahasan

Dengan menggunakan induksi matematika dimana Untuk n = 1 maka Untuk n = k maka di asumsikan terbukti Untuk n = k+1 maka jadi terbukti bahwa karena hasil sisi kanan dan kiri sama

Dengan menggunakan induksi matematika dimana

Untuk n = 1 maka

Untuk n = k maka di asumsikan terbukti

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 fraction numerator 2 straight k minus 1 over denominator 2 to the power of straight k minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2 straight n plus 3 over denominator 2 to the power of straight n minus 1 end exponent end fraction end cell row cell sum from straight i equals 1 to straight k plus 1 of fraction numerator 2 straight k minus 1 over denominator 2 to the power of straight k minus 1 end exponent end fraction end cell equals cell sum from straight i equals 1 to straight k of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction plus sum from straight i equals straight k plus 1 to straight k plus 1 of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i minus 1 end exponent end fraction end cell row cell 6 minus fraction numerator 2 left parenthesis straight k plus 1 right parenthesis plus 3 over denominator 2 to the power of straight k plus 1 minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2 straight k plus 3 over denominator 2 to the power of straight k minus 1 end exponent end fraction plus fraction numerator 2 left parenthesis straight k plus 1 right parenthesis minus 1 over denominator 2 to the power of straight k plus 1 minus 1 end exponent end fraction end cell row cell 6 minus fraction numerator 2 straight k plus 2 plus 3 over denominator 2 to the power of straight k end fraction end cell equals cell 6 minus fraction numerator 2 straight k plus 3 over denominator 2 to the power of straight k minus 1 end exponent end fraction plus fraction numerator 2 straight k plus 2 minus 1 over denominator 2 to the power of straight k end fraction end cell row cell 6 minus fraction numerator 2 straight k plus 5 over denominator 2 to the power of straight k end fraction end cell equals cell 6 minus fraction numerator 2 straight k plus 3 over denominator 2 to the power of straight k minus 1 end exponent end fraction plus fraction numerator 2 straight k plus 1 over denominator 2 to the power of straight k end fraction end cell row cell 6 minus fraction numerator 2 straight k plus 5 over denominator 2 to the power of straight k end fraction end cell equals cell 6 minus fraction numerator 1 plus 3 plus 2 straight k plus 1 over denominator 2 to the power of straight k end fraction end cell row cell 6 minus fraction numerator 2 straight k plus 5 over denominator 2 to the power of straight k end fraction end cell equals cell 6 minus fraction numerator 2 straight k plus 5 over denominator 2 to the power of straight k end fraction rightwards arrow Terbukti end cell end table$

jadi terbukti bahwa $sum from straight i equals 1 to straight n of fraction numerator 2 straight k minus 1 over denominator 2 to the power of straight k minus 1 end exponent end fraction equals 6 minus fraction numerator 2 straight n plus 3 over denominator 2 to the power of straight n minus 1 end exponent end fraction$ karena hasil sisi kanan dan kiri sama

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