Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
(→Solution 2) |
(→Solution 2) |
||
Line 78: | Line 78: | ||
<cmath>P(x)=(x-1)(x-2)(x-3)(x-m)+k</cmath> | <cmath>P(x)=(x-1)(x-2)(x-3)(x-m)+k</cmath> | ||
for some <math>m</math> and <math>f(1)=f(2)=f(3)=k</math>. Expanding gets | for some <math>m</math> and <math>f(1)=f(2)=f(3)=k</math>. Expanding gets | ||
− | |||
<cmath>P(x)=x^4-(m+6)x^3+(6m+11)x^2-(11m+6)x+6m+k</cmath> | <cmath>P(x)=x^4-(m+6)x^3+(6m+11)x^2-(11m+6)x+6m+k</cmath> | ||
Using the corresponding coefficient of <math>x^3</math> , we have | Using the corresponding coefficient of <math>x^3</math> , we have |
Revision as of 13:47, 29 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
Thus we obtain the telescoping series
Simplifying the sum we are left with
Thus .
~ Nafer
Let such that
for some
and
. Expanding gets
Using the corresponding coefficient of
, we have
Thus
By long division,
(
.
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 |