MathLabs

Problem 6

Let [x][x] denote the greatest integer not exceeding xx. If nn is a positive integer, express [n+12]+[n+24]+[n+48]+⋯[\frac{n+1}{2}]+[\frac{n+2}{4}]+[\frac{n+4}{8}]+\cdots as a function of nn.
Step 5 of 5: Take the tail to zero
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∑k=0∞[n+2k2k+1]=n\sum_{k=0}^{\infty}\left[\frac{n+2^k}{2^{k+1}}\right]=n
Detailed analysis

For 2N+1>n2^{N+1}>n, the last floor is 00, and all later summands are also 00. Therefore the requested sum equals nn.