How do you find the closed form of this summation?

$$\sum_{i=0}^{\log_4 n-1} i^2 $$

I know the following: $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$

How can I use this to find the closed form of my summation? Thanks

  • $\begingroup$ kindly check the first equation to see if $n^2$ is suppose to be $i^2$. $\endgroup$ – Siong Thye Goh Oct 21 '18 at 8:06
  • $\begingroup$ Thanks, I fixed it $\endgroup$ – kelp99 Oct 21 '18 at 8:09


Assuming $n$ is a power of $4$.

$$\sum_{i=0}^{\log_4 n-1} i^2 = \sum_{i=1}^{\log_4 n-1} i^2$$

Now, you can use the fomula that you listed in your question. Just replace $n$ in the formula by the relevant expression.


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.