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The sum of the first $n$ terms of the series
$1^{2} + 2 . 2^{2} + 3^{2} + 2 . 4^{2} + 5^{2} + 2 {.6}^{2} + ..........$ is $\frac{n {(n + 1)}^{2}}{2},$ when $n$ is even. When $n$ is odd the sum is

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n2(n+1)2

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n(n+1)2

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detailed solution

Correct option is A

When n is odd, last term will be n2 Then, the sum is 12+2 . 22+32+2 . 42+52+2 .62+..........+2(n−1)2+n2 =(n−1)n22+n2[Replacing n by n−1 in n(n+1)22] =n3−n2+2n22=n3+n22=n2(n+1)2

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The sum of the first n terms of the series 12+2 . 22+32+2 . 42+52+2 .62+.......... is n(n+1)22, when n is even. When n is odd the sum is

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Ready to Test Your Skills?

free study material

The sum of the first $n$ terms of the series
$1^{2} + 2 . 2^{2} + 3^{2} + 2 . 4^{2} + 5^{2} + 2 {.6}^{2} + ..........$ is $\frac{n {(n + 1)}^{2}}{2},$ when $n$ is even. When $n$ is odd the sum is