InterviewSolution
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InterviewSolution
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The sum to `n` terms of the series `3/(1^2)+5/(1^2+2^2)+7/(1^2+2^2+3^2)+-------` isA. `(6n)/(n+1)`B. `(9n)/(n+1)`C. `(12n)/(n+1)`D. `(3n)/(n+1)` |
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Answer» Correct Answer - A Let `T_(r)` be the `r^(th)` term of the given series. Then, `T_(r)=(2r+1)/(1^(2)+2^(2)+ . . . +r^(2))=(2r+1)/((r//6)(r+1)(2r+1))=6((1)/(r)-(1)/(r+1))` So, required sum is given by `underset(r=1)overset(n)sumT_(r)=6underset(r=1)overset(n)sum((1)/(r)-(1)/(r+1))` `rArr" "underset(r=1)overset(n)sum*T_(r)=6{(1-(1)/(2))+((1)/(2)-(1)/(3))+((1)/(3)-(1)/(4))+ . . . +((1)/(n)-(1)/(n+1))}` `rArr" "underset(r=1)overset(n)sumT_(r)=6{1-(1)/(n+1)}=(6n)/(n+1)` |
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