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Gunakan prinsip induksi matematika untuk membuktikan setiap pernyataan berikut b. P n ​ ≡ i = 1 ∑ n ​ a ix = a x − 1 a ( n + 1 ) x − a x ​ dengan x  = 0

Gunakan prinsip induksi matematika untuk membuktikan setiap pernyataan berikut 

b.   

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N. Puspita

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menggunakan pembuktian induksi matematika dimana untuk n = 1 akan dibuktikan Untuk n = k akan diasumsikan terbukti maka untuk n = k+1 akan dibuktikan jadi terbukti bahwa karena sisi kiri dan kanan sama

menggunakan pembuktian induksi matematika dimana untuk n = 1 akan dibuktikan

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to straight n of straight a to the power of ix end cell equals cell fraction numerator straight a to the power of open parentheses straight n plus 1 close parentheses straight x end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction space end cell row cell sum from straight i equals 1 to 1 of straight a to the power of straight x end cell equals cell fraction numerator straight a to the power of open parentheses 1 plus 1 close parentheses straight x end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction space end cell row cell straight a to the power of straight x end cell equals cell fraction numerator straight a to the power of 2 straight x end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction end cell row cell straight a to the power of straight x end cell equals cell fraction numerator straight a to the power of straight x left parenthesis straight a to the power of straight x minus 1 right parenthesis over denominator straight a to the power of straight x minus 1 end fraction end cell row cell straight a to the power of straight x end cell equals cell straight a to the power of straight x 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 i equals 1 to straight k of straight a to the power of ix end cell equals cell fraction numerator straight a to the power of open parentheses straight k plus 1 close parentheses straight x end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction space rightwards arrow terbuk ti end cell end table

untuk n = k+1 akan dibuktikan

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight i equals 1 to straight n of straight a to the power of ix end cell equals cell fraction numerator straight a to the power of open parentheses straight n plus 1 close parentheses straight x end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction space end cell row cell sum from straight i equals 1 to straight k plus 1 of straight a to the power of ix end cell equals cell sum from straight i equals 1 to straight k of straight a to the power of ix plus straight a to the power of left parenthesis straight k plus 1 right parenthesis straight x end exponent end cell row blank equals cell fraction numerator straight a to the power of open parentheses straight k plus 1 close parentheses straight x end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction plus space straight a to the power of left parenthesis straight k plus 1 right parenthesis straight x end exponent end cell row blank equals cell fraction numerator straight a to the power of open parentheses straight k plus 1 close parentheses straight x end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction minus fraction numerator straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction plus straight a to the power of open parentheses straight k plus 1 close parentheses straight x end exponent end cell row blank equals cell straight a to the power of open parentheses straight k plus 1 close parentheses straight x end exponent open parentheses fraction numerator 1 over denominator straight a to the power of straight x minus 1 end fraction plus 1 close parentheses minus fraction numerator straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction end cell row blank equals cell straight a to the power of open parentheses straight k plus 1 close parentheses straight x end exponent open parentheses fraction numerator 1 plus straight a to the power of straight x minus 1 over denominator straight a to the power of straight x minus 1 end fraction close parentheses minus fraction numerator straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction end cell row blank equals cell straight a to the power of open parentheses straight k plus 1 close parentheses straight x end exponent open parentheses fraction numerator straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction close parentheses minus fraction numerator straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction end cell row blank equals cell fraction numerator straight a to the power of open parentheses straight k plus 1 close parentheses straight x end exponent. straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction minus fraction numerator straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction end cell row blank equals cell fraction numerator straight a to the power of kx plus 2 straight x end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction end cell row blank equals cell fraction numerator straight a to the power of blank to the power of open square brackets left parenthesis straight k plus 1 right parenthesis plus 1 close square brackets straight x end exponent end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 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 straight a to the power of ix equals fraction numerator straight a to the power of open parentheses straight n plus 1 close parentheses straight x end exponent minus straight a to the power of straight x over denominator straight a to the power of straight x minus 1 end fraction space dengan space straight x not equal to 0 karena sisi kiri dan kanan sama

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