Problem:
If ⌊x⌋ is the greatest integer less than or equal to x, then
N=1∑1024⌊log2N⌋=
Answer Choices:
A. 8192
B. 8204
C. 9218
D. ⌊log2(1024!)⌋
E. none of these
Solution:
⌊log2N⌋=⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧12910 for for ⋅⋅ for for 2≤N<2222≤N<23⋅⋅29≤N<210N=210
Thus the desired sum is
1(22−2)+2(23−22)+3(24−23)+⋯+9(210−29)+10=9⋅210−(29+28+27+⋯+2)+10=9⋅210−(29+28+27+⋯+2+1)+11=9⋅210−(210−1)+11=8⋅210+12=8(1024)+12=8204
where we have used the sum formula for a geometric series to obtain the next to last line.