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Find the sum of a series $$\ \rm 1^2+2(2)^2+3^2+2(4)^2+5^2.......+2(n-1)^2+n^2$$ where n is odd

Attempt

I tried $$\ \rm \sum^n_1 [2(n-1)^2+n^2] - 2\sum^\frac{n-1}{2}_1 [2(n-1)^2+n^2] $$ but the answer is coming wrong.

The reasoning behind my procedure is that the expression $\ \rm \sum^n_1 [2(n-1)^2+n^2]$ will calculate the sum of all terms up til $n$ and the expression $\ \rm -2\sum^\frac{n-1}{2}_1 [2(n-1)^2+n^2]$ will erase the terms for which $n$ was even.

Can someone provide me some hint necessary for the correct answer and also provide a little insight towards the flaw in my working?

mathnoob123
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1 Answers1

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let $n = 2k+1$ for some integer $k$ then $k = \frac{n-1}{2}$ so your series becomes $$\sum_{k=0}^{\frac{n-1}{2}}(2k+1)^2 + 2(2k)^2$$

Pretty self explanatory after this.

Ziad Fakhoury
  • 2,593