Difference between revisions of "1995 AIME Problems/Problem 13"
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\left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4 &= \left(k^4 + 2k^3 + \frac 32k^2 + \frac 12k + \frac 1{16}\right) - \left(k^4 - 2k^3 + \frac 32k^2 - \frac 12k + \frac 1{16}\right)\\ &= 4k^3 + k. \end{align*}</cmath> | \left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4 &= \left(k^4 + 2k^3 + \frac 32k^2 + \frac 12k + \frac 1{16}\right) - \left(k^4 - 2k^3 + \frac 32k^2 - \frac 12k + \frac 1{16}\right)\\ &= 4k^3 + k. \end{align*}</cmath> | ||
− | Thus, <math>\frac{1}{k}</math> appears in the summation <math>4k^3 + k</math> times, and the sum for each <math>k</math> is then <math>(4k^3 + k) \cdot \frac{1}{k} = 4k^2 + 1</math>. From <math>k = 1</math> to <math>k = 6</math>, we get <math>\sum_{k=1}^{6} 4k^2 + 1 = 364 + 6 = | + | Thus, <math>\frac{1}{k}</math> appears in the summation <math>4k^3 + k</math> times, and the sum for each <math>k</math> is then <math>(4k^3 + k) \cdot \frac{1}{k} = 4k^2 + 1</math>. From <math>k = 1</math> to <math>k = 6</math>, we get <math>\sum_{k=1}^{6} 4k^2 + 1 = 364 + 6 = 370</math> (either adding or using the [[perfect square|sum of consecutive squares formula]]). |
− | But this only accounts for <math>\sum_{k = 1}^{6} (4k^3 + k) = 4\left(\frac{6(6+1)}{2}\right)^2 + \frac{6(6+1)}{2} = 1764 + 21 = 1785</math> terms, so we still have <math>1995 - 1785 = 210</math> terms with <math>f(n) = 7</math>. This adds <math>210 \cdot \frac {1}{7} = | + | But this only accounts for <math>\sum_{k = 1}^{6} (4k^3 + k) = 4\left(\frac{6(6+1)}{2}\right)^2 + \frac{6(6+1)}{2} = 1764 + 21 = 1785</math> terms, so we still have <math>1995 - 1785 = 210</math> terms with <math>f(n) = 7</math>. This adds <math>210 \cdot \frac {1}{7} = 30</math> to our summation, giving <math>{400}</math>. |
== See also == | == See also == |
Revision as of 00:17, 16 December 2014
Problem
Let be the integer closest to
Find
Solution
When , then
. Thus there are
values of
for which
. Expanding using the binomial theorem,
Thus, appears in the summation
times, and the sum for each
is then
. From
to
, we get
(either adding or using the sum of consecutive squares formula).
But this only accounts for terms, so we still have
terms with
. This adds
to our summation, giving
.
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.