InterviewSolution
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InterviewSolution
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The sum of the 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. 71B. 96C. 142D. 192 |
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Answer» Correct Answer - B `T_(n)=(1^3)+2^(3)+3^(3)+….+n^(3))/(1+3+5+…+(2n-1))=((n^(2)(n+1)^(2))/4)/(n/2(1+2n-1))` `rArrT_(n)=((n^(2)(n+1)^(2))/4)/(n^(2))` `thereforeT_(n)=1/4(n+1)^(2)` `T_(n)=1/4[n^(2)+2n+1]` `S_(n)=sum_(n=1)^(n)T_(n)` `S_(n)=1/4[(n(n+1)(2n+1))/6+n(n+1)+n]` Putting n=9, we get `S_(9)=1/4[(9xx10xx19)/6+9xx10+9]` `=1/4[285+90+9]=384/4=96` |
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