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**tony123****Member**- Registered: 2007-08-03
- Posts: 197

Evaluate the sum

.

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

I'll give it a crack. (For convenience I'll refer to the floor of the log as as log)

Assuming what you have in the sum is the floor function, we can do this in parts.

1. log1 = 0

2. log2+...+log3 = 2(1)

3. log4+...+log7 = (7-3)(2)

4. log8+...+log15 = (15-7)(3)

5. log16+...+log31 = (31-15)(4)

6. log32+...+log63 = (63-31)(5)

7. log64+...+log127 = (127-63)(6)

8. log128+...+log255 = (255-127)(7)

9. log256+...+log511 = (511-255)(8)

10. log512+...+log1023 = (1023-511)(9)

11. log1024+...+log2002 = (2002-1023)(10)

So the sum = 1(2) + 2(4)+3(8)+4(16)+5(32)+6(64)+7(128)+8(256)+9(512)+10(979) = 17984

Luckily we got in increase in the breadth of numbers the powers of 2 could handle, so it can be done this way. If you're smart you'll catch some sort of pattern early and it'll be a breeze to evaluate the sum. I'm not, so I took a while to see it.

Because I'm bored I'll express the sum in summation notation (i don't know how to do the last bit):

Now you might have some fancy formula for this, but using my calculator I get 17984.

*Last edited by Identity (2007-11-15 21:51:51)*

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