Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | {{ | + | Let <math>S = \sum_{k = 0}^{2005}\dfrac{k^2}{2^k}\cdot{2005 \choose k} = \sum_{k = 1}^{2005}\dfrac{k^2}{2^k}\cdot{2005 \choose k}</math>. Let <math>W = \left(\dfrac{2}{3}\right)^{2005}S</math>. Then note that <math>(x + 1)^{2005} = \sum_{k = 0}^{2005} {2005 \choose k}x^k</math>, so taking the derivative and multiplying by <math>x</math> gives <math>2005x(x + 1)^{2004} = \sum_{k = 0}^{2005} k{2005 \choose k}x^k</math>. Taking the derivative and multiplying by <math>x</math> again gives <math>f(x) = 2005x(x + 1)^{2004} + (2005)(2004)x^2(x + 1)^{2003} = \sum_{k = 0}^{2005} k^2{2005 \choose k}x^k</math>. Now note that <math>f\left(\dfrac{1}{2}\right) = S</math>. Then we get <math>W = \left(\dfrac{2}{3}\right)^{2005}S = \left(\dfrac{2}{3}\right)^{2005}\left(\dfrac{2005}{2}\left(\dfrac{3}{2}\right)^{2004} + (501)(2005)\left(\dfrac{3}{2}\right)^{2003}\right)</math>, so <math>W = \dfrac{2}{3}\cdot\dfrac{2005}{2} + \dfrac{668}{3}\cdot(2005) = 447115</math>, so <math>W \equiv \boxed{115} \pmod{1000}</math>. |
==See Also== | ==See Also== | ||
{{Mock AIME box|year=Pre 2005|n=3|num-b=12|num-a=14}} | {{Mock AIME box|year=Pre 2005|n=3|num-b=12|num-a=14}} | ||
Revision as of 00:10, 25 April 2013
Problem
Let 
denote the value of the sum
Determine the remainder obtained when is divided by 
.
Solution
Let 
. Let 
. Then note that 
, so taking the derivative and multiplying by 
gives 
. Taking the derivative and multiplying by again gives 
. Now note that 
. Then we get 
, so 
, so 
.
See Also
| Mock AIME 3 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
