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The sum of first $n$ terms of the given series $1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+$\ldots\ldots\ldots is $\frac{n(n+1)^2}{2}$, when $n$ is even. When $n$ is odd, the sum will be
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Verified Answer
The correct answer is:
$\frac{1}{2} n^2(n+1)$
When $n$ is odd, the last term. i.e., the $n^{\text {th }}$ term will be $n^2$ in this case $n-1$ is even and so the sum of the first
$n-1$ terms of the series is obtained by replacing $n$ by $n-1$ in the given formula and so is $\frac{1}{2}(n-1) n^2$ Hence the sum of the $n$ terms $=$ (the sum of $n-1$ terms) + the $n^{\text {th }}$ term $=\frac{1}{2}(n-1) n^2+n^2=\frac{1}{2}(n+1) n^2$
Trick : Check for $n=1,3$. Here $S_1=1, S_3=18$ which gives (2).
$n-1$ terms of the series is obtained by replacing $n$ by $n-1$ in the given formula and so is $\frac{1}{2}(n-1) n^2$ Hence the sum of the $n$ terms $=$ (the sum of $n-1$ terms) + the $n^{\text {th }}$ term $=\frac{1}{2}(n-1) n^2+n^2=\frac{1}{2}(n+1) n^2$
Trick : Check for $n=1,3$. Here $S_1=1, S_3=18$ which gives (2).
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