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Series sum

Source: AHSME 1986 problem 25

October 2, 2011
floor functionlogarithmsAMC

Problem Statement

If ⌊x⌋\lfloor x\rfloor is the greatest integer less than or equal to xx, then ∑N=11024⌊log⁡2N⌋=\displaystyle\sum_{N=1}^{1024} \lfloor \log_{2}N\rfloor =
<spanclass=′latex−bold′>(A)</span> 8192<spanclass=′latex−bold′>(B)</span> 8204<spanclass=′latex−bold′>(C)</span> 9218<spanclass=′latex−bold′>(D)</span> ⌊log⁡2(1024!)⌋<spanclass=′latex−bold′>(E)</span> none of these <span class='latex-bold'>(A)</span>\ 8192\qquad<span class='latex-bold'>(B)</span>\ 8204\qquad<span class='latex-bold'>(C)</span>\ 9218\qquad<span class='latex-bold'>(D)</span>\ \lfloor \log_{2}(1024!)\rfloor\qquad<span class='latex-bold'>(E)</span>\ \text{none of these}
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