InterviewSolution
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For the series, `S=1 +(1)/(1+3)(1+2)^2+(1)/((1+3+7))(1+2+3)^2+(1)/((1+3+5+7))(1+2+3+4)^2+...`A. `7^(th) " term is " 16`B. `7^(th) " term " is 18`C. Sum of first 10 terms is `(505)/(4)`D. Sum of first 10 terms is `(405)/(4)` |
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Answer» Correct Answer - A::C `S=1+1/((1+3))(1+2)^(2)+1/((1+3+5))(1+2+3)^(2)+1/((1+3+5+7))(1+2+3+4)^(2)+…` The `r^(th)` term is given by `T_(r)=1/r^(2)(1+2+..+r)^(2)=1/r^(2){(r(r+1))/2}^(2)=(r^(2)+2r+1)/4` `thereforeT_(7)=16` and `S_(10)=sum_(r=1)^(10)T_(r)` =`1/4{((10)(10+1)(20+1))/6+(10)(10+1)+10}` `=505/4` |
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