A proving question based on greatest integer function.

Question

If $$[x]=[\frac{x}{2}]+[\frac{x+1}{2}] $$b , where [°] denotes the greatest integer function and $n$ be a positive integer, then show that $$[\frac{n+1}{2}]+[\frac{n+2}{4}]+[\frac{n+4}{8}]+[\frac{n+8}{16}]\cdots = n$$. I tried to solve this problem by removing the integer parts and then adding the left over parts but could not get the result please help me out.

Answer 1

$$[n]=[\frac{n}{2}]+[\frac{n+1}{2}]$$ $$[\frac{n}{2}]=[\frac{n}{4}]+[\frac{\frac{n}{2}+1}{2}]=[\frac{n}{4}]+[\frac{n+2}{4}]$$ $$[\frac{n}{4}]=[\frac{n}{8}]+[\frac{\frac{n}{4}+1}{2}]=[\frac{n}{8}]+[\frac{n+4}{8}]$$ and so on. Use telescopic sums to get the answer ;).