Question
The value of $(n-2)^{2}+(n-4)^{2}+(n-6)^{2}+\ldots$ to $n$ terms is(A) $\frac{n}{3}\left(n^{2}+2\right)$(B) $\frac{n}{2}\left(n^{2}+3\right)$(C) $\frac{n}{3}\left(n^{2}-2\right)$(D) $\frac{n}{2}\left(n^{2}-3\right)$
Step 1
The nth term of the sequence can be written as $(n-2i)^2$ where $i$ ranges from 0 to $n-1$. Show more…
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