Roboguru

Gunakan prinsip induksi matematika untuk membuktikan setiap pernyataan berikut  a.

Pertanyaan

Gunakan prinsip induksi matematika untuk membuktikan setiap pernyataan berikut 

a.  straight P subscript straight n identical to sum from straight i equals 1 to straight n of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i 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    

Pembahasan Soal:

Dengan menggunakan 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 fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i 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 1 of fraction numerator 2.1 minus 1 over denominator 2 to the power of 1 minus 1 end exponent end fraction end cell equals cell 6 minus fraction numerator 2.1 plus 3 over denominator 2 to the power of 1 minus 1 end exponent end fraction end cell row cell fraction numerator 2 minus 1 over denominator 2 to the power of 0 end fraction end cell equals cell 6 minus fraction numerator 2 plus 3 over denominator 2 to the power of 0 end fraction end cell row cell 1 over 1 end cell equals cell 6 minus 5 over 1 end cell row 1 equals cell 1 rightwards arrow Terbukti end cell end table

Untuk n = k maka di asumsikan terbukti

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 i minus 1 over denominator 2 to the power of straight i 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 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 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 end cell row cell fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i 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 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 i equals 1 to straight n 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 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 i minus 1 over denominator 2 to the power of straight i 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 straight P subscript straight n identical to sum from straight i equals 1 to straight n of fraction numerator 2 straight i minus 1 over denominator 2 to the power of straight i 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

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 18 Juli 2021

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