If the sum of the series 1^2+22^2+3^2+24^2+5^2+26^2+ up to n terms, when n is even, is n(n…

Question

If the sum of the series $1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+2\cdot6^2+\cdots$ up to $n$ terms, when $n$ is even, is $\dfrac{n(n+1)^2}{2}$, then the sum of the series, when $n$ is odd, is:

Accepted Answer

For odd $n$, the sum is the even-$n$ sum through $n-1$ terms plus the final term $n^2$ (unpaired, since $n$ is odd its own term is not doubled):

$$S(n)=S(n-1)+n^2=\frac{(n-1)\,n^2}{2}+n^2=n^2\left(\frac{n-1}{2}+1\right)=\frac{n^2(n+1)}{2}.$$

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