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Prove by induction that k = 1 ∑ n ​ ( k + 1 ) ⋅ 2 k − 1 = n ⋅ 2 n

Prove by induction that

 

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A. Acfreelance

Master Teacher

Mahasiswa/Alumni UIN Walisongo Semarang

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terbukti untuk karena hasil dari sisi kiri dan kanan sama

terbukti untuk sum from straight k equals 1 to straight n of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent equals straight n times 2 to the power of straight n karena hasil dari sisi kiri dan kanan sama

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Pembahasan

Untuk n = 1 maka Untuk n = k diasumsikan terbukti maka Untuk n = k+1 maka Jadi terbukti untuk karena hasil dari sisi kiri dan kanan sama

Untuk n = 1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight k equals 1 to straight n of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent end cell equals cell straight n times 2 to the power of straight n end cell row cell sum from straight k equals 1 to 1 of open parentheses 1 plus 1 close parentheses times 2 to the power of 1 minus 1 end exponent end cell equals cell 1 times 2 to the power of 1 end cell row cell 2.1 end cell equals cell 1.2 end cell row 2 equals cell 2 rightwards arrow Terbukti space 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 k equals 1 to straight n of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent end cell equals cell straight n times 2 to the power of straight n end cell row cell sum from straight k equals 1 to straight k of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent end cell equals cell straight k times 2 to the power of straight k end cell row blank equals cell 2 straight k to the power of straight k 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 k equals 1 to straight n of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent end cell equals cell straight n times 2 to the power of straight n end cell row cell sum from straight k equals 1 to straight k of open parentheses straight k plus 1 plus 1 close parentheses times 2 to the power of straight k plus 1 minus 1 end exponent end cell equals cell left parenthesis straight k plus 1 right parenthesis times 2 to the power of left parenthesis straight k plus 1 right parenthesis end exponent end cell row cell open parentheses straight k plus 2 close parentheses.2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 2 to the power of straight k plus 1 end exponent end cell row cell 2 straight k to the power of straight k plus 4 to the power of straight k plus 2 straight k to the power of straight k end cell equals cell 2 straight k to the power of straight k.2 straight k to the power of 1 plus 2 to the power of straight k.2 to the power of 1 end cell row cell 4 straight k to the power of straight k plus 4 to the power of straight k end cell equals cell 4 straight k to the power of straight k plus 4 to the power of straight k rightwards arrow Terbukti end cell end table

Jadi terbukti untuk sum from straight k equals 1 to straight n of open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 end exponent equals straight n times 2 to the power of straight n karena hasil dari sisi kiri dan kanan sama

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