# Summation involving floor function with powers

I want to perform $\sum\limits_{i=1}^k \lfloor\frac{N}{2^i}\rfloor7^{i-1}$ where $N$ is an even integer.

Is there any technique which performs it?

THanks.

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Actually, now that I think about it, the edit does not make much sense since the powers of $2's$ cancel to give $\lfloor N/2 \rfloor$. Can the OP please clarify? Thanks. –  Sandeep Silwal Apr 14 '14 at 5:01
Note that the powers of 2 almost cancel out against each other. –  Marc Apr 14 '14 at 5:03
What is $N$? Integer, real? If $N$ is an integer, it is helpful to write $N = 2^ab$ with $b$ odd. –  Sandeep Silwal Apr 14 '14 at 7:18
Yea N is an even integer! –  user1234 Apr 14 '14 at 11:19