InterviewSolution
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InterviewSolution
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Show that S_(n)=(n(2n^(2)+9n+13))/(24). |
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Answer» Solution :Let `T_(N)` and `T_(n)'` be the nth terms of the series in numerator and denominaror of LHS. Then, `THEREFORE T_(n)=n(n+1)^(2)" and " T_(n)'=n^(2)(n+1)` `therefore LHS=(sumT_(n))/(sumT_(n))=(sumn(n+1)^(2))/(sumn^(2)(n+1))=(sum(n^(3)+2N^(2)+n))/(sum(n^(3)+n^(2)))` `=(sumn^(3)+2sumn^(2)+sumn)/(sumn^(3)+sumn^(2))` `=({n(n+1)/(2)}^(2)+2{n(n+1)(2n+1)/(6)}+{n(n+1)/(2)})/({n(n+1)/(2)}^(2)+{(n(+1)(2n+1))/(6)})` `=((n(n+1))/(2){(n(n+1))/(2)+(2(2n+1))/(3)+1})/((n(n+1))/(2){(n(n+1))/(2)+(2n+1)/(3)})` `=((1)/(6)(3n^(2)+3n+8n+4+6))/((1)/(6)(3n^(2)+3n+4n+2))` `=((3n^(2)+11n+10))/((3n^(2)+7n+2))=((3n+5)(n+2))/((3n+1)(n+2))=((3n+5))/((3n+1))=RHS`. |
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