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Find the sum of the series `1^2+3^2+5^2+ ton`terms. |
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Answer» Correct Answer - `(n(2n-1)(2n+1))/(3)` `1^(2)+3^(2)+5^(2)+…=Sigma(2n-1)^(2)` `=Sigma(4n^(2)-4n+1)` `=4Sigman^(2)-4Sigman+n` `=4(n(n+1)(2n+1))/3-2n(n+1)+n` `=n/3(4n^(2)-1)` `=(n(2n-1)(2n+1))/3` |
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