Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
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<cmath>2\cdot(n+1)\cdot b^2c=n^2-1</cmath> | <cmath>2\cdot(n+1)\cdot b^2c=n^2-1</cmath> | ||
<cmath>b^2c=\frac{(n-1)(n+1)}{2\cdot(n+1)}</cmath> | <cmath>b^2c=\frac{(n-1)(n+1)}{2\cdot(n+1)}</cmath> | ||
+ | <cmath>b\sqrt{c}=\sqrt{\frac{n-1}{2}}</cmath> | ||
==See Also== | ==See Also== |
Revision as of 14:19, 28 November 2019
Contents
[hide]Problem
Let denote the value of the sum
can be expressed as
, where
and
are positive integers and
is not divisible by the square of any prime. Determine
.
Solution
Notice that . Thus, we have
This is a telescoping series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with
, and
.
Solution 2
Simplifying the expression yields
Now we can assume that
for some
,
,
.
Squaring the first equation yields
which gives the system of equations
calling them equations
and
, respectively.
Also we have
which obtains equation
.
Adding equations and
yields
Squaring equation
and substituting yields
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |