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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

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  • $\begingroup$ 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
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Guide:

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.

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