https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Phil9047&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-23T19:26:24Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2008_Mock_ARML_2_Problems/Problem_8&diff=78756 2008 Mock ARML 2 Problems/Problem 8 2016-05-31T04:57:32Z <p>Phil9047: /* Solution */</p> <hr /> <div>==Problem==<br /> Given that &lt;math&gt;\sum_{i = 0}^{n}a_ia_{n - i} = 1&lt;/math&gt; and &lt;math&gt;a_n &gt; 0&lt;/math&gt; for all non-negative integers &lt;math&gt;n&lt;/math&gt;, evaluate &lt;math&gt;\sum_{j = 0}^{\infty}\frac {a_j}{2^j}&lt;/math&gt;.<br /> <br /> ==Solution==<br /> 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]]). <br /> <br /> Let &lt;math&gt;x&lt;/math&gt; be an arbitrary number; note that <br /> &lt;center&gt;&lt;math&gt;\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&lt;/math&gt;&lt;/center&gt;<br /> By the given, the coefficients on the right-hand side are all equal to &lt;math&gt;1&lt;/math&gt;, yielding the [[geometric series]]:<br /> &lt;center&gt;&lt;math&gt;\left[\sum_{j = 0}^{\infty} a_jx^j\right]^2 = 1 + x + x^2 + \cdots = \frac{1}{1-x}&lt;/math&gt;&lt;/center&gt;<br /> For &lt;math&gt;x = \frac{1}{2}&lt;/math&gt;, this becomes &lt;math&gt;\left[\sum_{j = 0}^{\infty}\frac {a_j}{2^j}\right]^2 = 2&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{\sqrt{2}}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{Mock ARML box|year = 2008|n = 2|num-b=7|after=Final Question|source = 206880}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Phil9047