1995 AIME Problems/Problem 13
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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 |
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