InterviewSolution
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InterviewSolution
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Find the sum `1^2+(1^2+2^2)+(1^2+2^2+3^2)+`up to 22nd term. |
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Answer» Correct Answer - 23276 `T_(n)=1^(2)+2^(2)+3^(2)+…+n^(2)` `=(n(n+1)(2n+1))/6` `=n/6(2n^(2)+3n+1)` `=(2n^(3)+3n^(2)+n)/6` `S_(n)=sumT_(n)` `=1/6[2sumn^(3)+3sumn^(2)+sumn]` `=1/6{2[(n(n+1))/2]^(2)+3xx(n(n+1)(2n+1))/6+(n(n+1))/2}` `=1/6[(n^(2)(n+1)^(2))/2+(n(n+1)(2n+1))/6+(n(n+1))/2]` `=1/6xx1/2n(n+1)[n(n+1)[n(n+1)+(2n+1)+1]` `=(n(n+1))/12[n^(2)+3n+2]` For n= 22 `S_(22)=(22xx23)/12[22^(2)+66+2]` `=(22xx23)12[552]` `=22xx23xx46` =23276 |
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