https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_2007-2008_Problems/Problem_2&feed=atom&action=historyMock AIME 1 2007-2008 Problems/Problem 2 - Revision history2024-03-28T20:14:25ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_2007-2008_Problems/Problem_2&diff=24470&oldid=prevAzjps: solution2008-04-02T21:25:08Z<p>solution</p>
<p><b>New page</b></p><div>== Problem ==<br />
The [[binomial expansion|expansion]] of <math>(x+1)^n</math> has 3 consecutive terms with [[coefficient]]s in the ratio <math>1:2:3</math> that can be written in the form <cmath>{n\choose k} : {n\choose k+1} : {n \choose k+2}</cmath><br />
Find the sum of all possible values of <math>n+k</math>. <br />
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== Solution ==<br />
By definition, <math>{n\choose k} = \frac{n!}{k!(n-k)!}</math>. The ratio of the first two terms give us that <cmath>\begin{align*}\frac{1}{2} &= \frac{\frac{n!}{k!(n-k)!}}{\frac{n!}{(k+1)!(n-k-1)!}} = \frac{k+1}{n-k}\\ 2&=n-3k\end{align*}</cmath>The ratio of the second and third terms give us that <cmath>\begin{align*}\frac{2}{3} &= \frac{\frac{n!}{(k+1)!(n-k-1)!}}{\frac{n!}{(k+2)!(n-k-2)!}} = \frac{k+2}{n-k-1}\\ 8&=2n-5k\end{align*}</cmath><br />
This is a linear system of two [[equation]]s with two unknowns, indicating that there is a unique solution. Solving by substitution or multiplying the top equation and subtracting, we find <math>k = 4, n = 14</math>. Thus, <math>n+k=\boxed{018}</math>.<br />
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== See also ==<br />
{{Mock AIME box|year=2007-2008|n=1|num-b=1|num-a=3|source=196980}}<br />
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[[Category:Intermediate Algebra Problems]]</div>Azjps