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?


2 Answers 2


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


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 $${50(101)33}=166650$$

  • 11
    $\begingroup$ Given the givens, you really don't need a closed forms - this question is about taking the givens and working out the third value. $\endgroup$ Mar 14, 2017 at 16:21
  • 1
    $\begingroup$ 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, 2017 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.