InterviewSolution
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InterviewSolution
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Find the sum of the series `(1^3)/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+3+5)+`up to `n`terms. |
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Answer» Let `T_(n)` be the nth term of the given series. Then, `T_(n)=(1^(3)+2^(3)+3^(3)+"..."n^(3))/((1+3+5+"...."+(2n-1)))={(n(n+1))/(2)}^(2)/((n)/(2)(1+2n-1))` `((n+1)^(2))/(4)=(1)/(4)(n^(2)+2n+1)` Let `S_(n)` denotes the sum of n terms of the given series. Then, `S_(n)=sumT_(n)=(1)/(4)sum(n^(2)+2n+1)` `=(1)/(4)(sumn^(2)+sum2n+sum1)` `=(1)/(4){n(n+1)(2n+1))/(6)+(2n(n+1))/(2)+n}` `=(n)/(24){2n^(2)+3n+1+6n+6+6}` Hence, `S_(n)=(n(2n^(2)+9n+13))/(24)` |
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