Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
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&= \sum_{n=1}^{9800}\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}} \\ | &= \sum_{n=1}^{9800}\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}} \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
+ | Simplifying the sum we are left with | ||
+ | <cmath>\begin{align*} | ||
+ | S &= -\sqrt{\frac{1}{2}}+\sqrt{\frac{9800}{2}}+\sqrt{\frac{9801}{2}} \\ | ||
+ | &= -\frac{\sqrt{2}}{2}+\frac{99\sqrt{2}}{2}+70 \\ | ||
+ | &= 70+49\sqrt{2} \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Thus <math>p+q+r=70+49+2=\boxed{121}</math>. | ||
+ | |||
+ | |||
+ | ~ Nafer | ||
==See Also== | ==See Also== |
Latest revision as of 14:48, 29 November 2019
Contents
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
Thus we obtain the telescoping series
Simplifying the sum we are left with
Thus .
~ Nafer
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 |