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Buktikan masing-masing notasi sigma berikut. d. p = 1 ∑ n ​ p ⋅ 2 p − 1 = ( n − 1 ) 2 n + 1

Buktikan masing-masing notasi sigma berikut.

d. 

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A. Acfreelance

Master Teacher

Mahasiswa/Alumni UIN Walisongo Semarang

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di buktikan menggunakan induksi matematika terbukti karena hasil ruas kanan dan kiri sama

sum from straight p equals 1 to straight n of straight p times 2 to the power of straight p minus 1 end exponent equals open parentheses straight n minus 1 close parentheses 2 to the power of straight n plus 1 di buktikan menggunakan induksi matematika terbukti karena hasil ruas kanan dan kiri sama

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Pembahasan

Buktikan dengan menggunkan induksi matematika dimana n = 1 maka n = k dimana diasumsikan terbukti maka Akan dibuktikan untuk n = k+1 maka PRK Jadi di buktikan menggunakan induksi matematika terbukti karena hasil ruas kanan dan kiri sama

Buktikan dengan menggunkan induksi matematika dimana

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 straight p times 2 to the power of straight p minus 1 end exponent end cell equals cell open parentheses straight n minus 1 close parentheses 2 to the power of straight n plus 1 end cell row cell sum from straight p equals 1 to 1 of straight p times 2 to the power of 1 minus 1 end exponent end cell equals cell open parentheses 1 minus 1 close parentheses 2 to the power of 1 plus 1 end cell row cell 1.2 to the power of 0 end cell equals cell 0.2 plus 1 end cell row 1 equals cell 1 rightwards arrow Terbukti end cell end table

n = k dimana diasumsikan terbukti maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight p equals 1 to straight n of straight p times 2 to the power of straight p minus 1 end exponent end cell equals cell open parentheses straight n minus 1 close parentheses 2 to the power of straight n plus 1 end cell row cell sum from straight p equals 1 to straight k of straight p times 2 to the power of straight p minus 1 end exponent end cell equals cell open parentheses straight k minus 1 close parentheses 2 to the power of straight k plus 1 end cell row blank equals cell 2 straight k to the power of straight k minus 2 to the power of straight k plus 1 rightwards arrow Terbukti end cell end table

Akan dibuktikan 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 straight p times 2 to the power of straight p minus 1 end exponent end cell equals cell open parentheses straight n minus 1 close parentheses 2 to the power of straight n plus 1 end cell row cell sum from straight p equals 1 to straight k plus 1 of straight p times 2 to the power of straight p minus 1 end exponent end cell equals cell open parentheses straight k plus 1 minus 1 close parentheses 2 to the power of left parenthesis straight k plus 1 right parenthesis end exponent plus 1 end cell row blank equals cell k.2 to the power of straight k plus 1 end exponent plus 1 end cell row blank equals cell 2 k to the power of straight k plus 1 end exponent plus 1 rightwards arrow Terbukti end cell end table

PRK

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight p equals 1 to straight k plus 1 of straight p times 2 to the power of straight p minus 1 end exponent end cell equals cell sum from straight p equals 1 to straight k of straight p times 2 to the power of straight p minus 1 end exponent plus sum from straight p equals straight k plus 1 to straight k plus 1 of straight p times 2 to the power of straight p minus 1 end exponent end cell row cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell equals cell 2 straight k to the power of straight k minus 2 to the power of straight k plus 1 plus open parentheses straight k plus 1 close parentheses.2 to the power of straight k plus 1 minus 1 end exponent end cell row cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell equals cell 2 straight k to the power of straight k minus 2 to the power of straight k plus 1 plus 2 to the power of straight k plus 2 straight k end cell row cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell equals cell 2 straight k to the power of straight k plus 2 straight k plus 1 end cell row cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 rightwards arrow terbukti end cell end table

Jadi sum from straight p equals 1 to straight n of straight p times 2 to the power of straight p minus 1 end exponent equals open parentheses straight n minus 1 close parentheses 2 to the power of straight n plus 1 di buktikan menggunakan induksi matematika terbukti karena hasil ruas kanan dan kiri sama

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