How to find closed form of summation

Question

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

Answer 1

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.