Difference between revisions of "2008 Mock ARML 2 Problems/Problem 8"
(New page: ==Problem== Given that <math>\sum_{i = 0}^{n}a_ia_{n - i} = 1</math> and <math>a_n > 0</math> for all non-negative integers <math>n</math>, evaluate <math>\sum_{j = 0}^{\infty}\frac {a_j}{...) |
m (→Solution) |
||
(5 intermediate revisions by 4 users not shown) | |||
Line 3: | Line 3: | ||
==Solution== | ==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 polynomials (or [[generating functions]]). | |
− | <math>a_0a_1+a_1a_0=1\ | + | Let <math>x</math> be an arbitrary number; note that |
+ | <center><math>\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</math></center> | ||
+ | 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> | ||
+ | For <math>x = \frac{1}{2}</math>, this becomes <math>\left[\sum_{j = 0}^{\infty}\frac {a_j}{2^j}\right]^2 = 2</math>, and the answer is <math>\boxed{\sqrt{2}}</math>. | ||
− | + | ==See also== | |
− | + | {{Mock ARML box|year = 2008|n = 2|num-b=7|after=Final Question|source = 206880}} | |
− | |||
− | |||
− | |||
− | + | [[Category:Intermediate Algebra Problems]] | |
− | |||
− | |||
− | |||
− | |||
− | |||
− |
Latest revision as of 23:57, 30 May 2016
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 polynomials (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 |