The sum to 50 terms of the series `3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+� – InterviewSolution

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1.	The sum to 50 terms of the series `3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+….+… is `A. `100/17`B. `150/17`C. `200/51`D. `50/17`
Answer» Correct Answer - A Let `T_(r)` be the rth term of the given series. Then, `T_(r)=(2r+1)/(1^(2)+2^(2)+…+r^(2))=(6(2r+1))/((r )(r+1)(2r+1))` `=6(1/r-1/(r+1))` So, sum is given by `sum_(r=1)^(50)T_(r)=6sum_(r=1)^(50)(1/r-1/(r+1))` `=6[(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/50-1/51)]` `=6[1-1/51]=100/17`

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