InterviewSolution
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InterviewSolution
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The sum of the first n terms of the series `1^2+ 2.2^2+3^2 +2.4^2+....` is `(n(n+1)^2)/2` when n is even. Then the sum if n is odd , is |
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Answer» Correct Answer - `(n^2(n+1))/(2)` When n is even, last term of the series is `n^(2)`. So, we have `1^(2)+2xx2^(2)+3^(2)+2xx4^(2)+5^(2)+2xx6^(2)+..+(n-1)^(2)+2xxn^(2)` `=(n(n+1)^(2))/2` ..(1) When n is odd, the series is `1^(2)+2xx2^(2)+3^(2)+2xx4^(2)+5^(2)+2xx6^(2)+....+2(n-1)^(2)+n^(2)` `therefore` Sum of series=`[1^(2)+2xx2^(2)+3^(2)+2xx4^(2)+...+2xx(n-1)^(2)]+n^(2)` `=((n-1)n^(2))/2+n^(2)` [In (1) replacing n by (n-1)] `=(n^(2)(n+1))/2` |
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