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We know that $1^1+2^2+...+100^2=338350$ and $1^1+2^2+...+50^2=42925$. Find $1^2+3^2+...99^2$.
I don't know really where to start. I tried to find a pattern in the sequences, but there was none. Can I substitute values for the equations?

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This should help: $$2^2+4^2+6^2+\dots+100^2=2^2(1^2+2^2+3^2+\dots+50^2)=\dots$$

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As mentioned in the comments the sum of the series can be derived in the following way

$$\sum_{i=1}^{2n}i^2=1^2+2^2+3^2+...+(2n)^2=\frac{2n(2n+1)(4n+1)}{6}$$ Similarly, $$\sum_{i=1}^{n}{(2i)}^2=2^2+4^2+...+(2n)^2=4\sum_{i=1}^{n}{i}^2=\frac{2n(n+1)(2n+1)}{3}$$

Subtacting gives required sum $$\sum_{i=1}^{n}{(2i-1)}^2=1^2+3^2+...+(2n-1)^2=\frac{2n(2n+1)(4n+1)}{6}-\frac{4n(n+1)(2n+1)}{6}=\frac{2n(2n+1)(4n+1-2n-2)}{6}=\frac{n(2n+1)(2n-1)}{3}$$

Put $n=50$ to get required solution. which gives $${50(101)33}=166650$$

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    $\begingroup$ Given the givens, you really don't need a closed forms - this question is about taking the givens and working out the third value. $\endgroup$ – Thomas Andrews Mar 14 '17 at 16:21
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    $\begingroup$ Well..It never hurts to know a little bit more...I presented this generalised solution so that it may help OP in the future ... $\endgroup$ – user35508 Mar 14 '17 at 16:25

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