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The sum of first 9 terms of the series (...

The sum of first 9 terms of the series `(1^(3))/(1)+(1^(3)+2^(3))/(1+3)+(1^(3)+2^(3)+3^(3))/(1+3+5)+"........"` is

A

192

B

71

C

96

D

142

Text Solution

Verified by Experts

The correct Answer is:
C
`T_n = (1^2 +2^2 +3^3 +...+n^3)/(1+3+5+...+(2n-1))=((n(n+1))/2)^2/((n(1+2n-1))/2)=((n+1)^2)/4`
` 1/4 (n^2+2n+1)`
`:." "S_n = 1/4(Sigman^2 +2Sigman+Sigma1)=1/4[(n(n+1)(2n+1))/6+(2n(n+1))/2+n]`
`S_9 = 1/4 [285 +90 +9] =96`
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