Difference between revisions of "2008 Mock ARML 2 Problems/Problem 8"
(solution) |
(→Solution) |
||
Line 6: | Line 6: | ||
Let <math>x</math> be an arbitrary number; note that | Let <math>x</math> be an arbitrary number; note that | ||
− | <center><math>\begin{align*}\left[\sum_{j = 0}^{\infty} a_jx^j\right]^2 &= (a_0 + a_1 \cdot x + a_2 \cdot x^2 + \cdots)^2\\ &= a_0^2 + (a_0a_1 + a_1a_0)x + (a_0a_2 + | + | <center><math>\begin{align*}\left[\sum_{j = 0}^{\infty} a_jx^j\right]^2 &= (a_0 + a_1 \cdot x + a_2 \cdot x^2 + \cdots)^2\\ &= a_0^2 + (a_0a_1 + a_1a_0)x + (a_0a_2 + a_1a_1 + a_2a_0)x^2 + \cdots\end{align*}</math></center> |
By the given, the coefficients on the right-hand side are all equal to <math>1</math>, yielding the [[geometric series]]: | By the given, the coefficients on the right-hand side are all equal to <math>1</math>, yielding the [[geometric series]]: | ||
<center><math>\left[\sum_{j = 0}^{\infty} a_jx^j\right]^2 = 1 + x + x^2 + \cdots = \frac{1}{1-x}</math></center> | <center><math>\left[\sum_{j = 0}^{\infty} a_jx^j\right]^2 = 1 + x + x^2 + \cdots = \frac{1}{1-x}</math></center> |
Revision as of 14:49, 2 June 2011
Problem
Given that and for all non-negative integers , evaluate .
Solution
The motivating factor for this solution is the form of the first summation, which might remind us of the expansion of the coefficients of the product of two multinomials (or generating functions).
Let be an arbitrary number; note that
By the given, the coefficients on the right-hand side are all equal to , yielding the geometric series:
For , this becomes , and the answer is .
See also
2008 Mock ARML 2 (Problems, Source) | ||
Preceded by Problem 7 |
Followed by Final Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 |