Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
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+ | Let <math>P(x)=x^4-3x^3-ax^2+bx-12</math> such that | ||
+ | <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 | ||
+ | <cmath>P(x)=(x^3−6x^2+11x−6)(x-m)+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 | ||
+ | <cmath>m+6=3\Longrightarrow m=-3</cmath> | ||
+ | Thus | ||
+ | <cmath>P(x)=x^4-3x^3-7x^2+27^x-12</cmath> | ||
+ | By long division, <math>f(x)=48</math> (<math>\boxed{C})</math>. | ||
==See Also== | ==See Also== |
Revision as of 13:45, 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
\[P(x)=(x^3−6x^2+11x−6)(x-m)+k\] (Error compiling LaTeX. Unknown error_msg)
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 |