Mock AIME 1 2007-2008 Problems/Problem 2

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Problem

The expansion of $(x+1)^n$ has 3 consecutive terms with coefficients in the ratio $1:2:3$ that can be written in the form ${n\choose k} : {n\choose k+1} : {n \choose k+2}$ Find the sum of all possible values of $n+k$ .

Solution

By definition, ${n\choose k} = \frac{n!}{k!(n-k)!}$ . The ratio of the first two terms give us that $\begin{align*}\frac{1}{2} &= \frac{\frac{n!}{k!(n-k)!}}{\frac{n!}{(k+1)!(n-k-1)!}} = \frac{k+1}{n-k}\\ 2&=n-3k\end{align*}$ The ratio of the second and third terms give us that $\begin{align*}\frac{2}{3} &= \frac{\frac{n!}{(k+1)!(n-k-1)!}}{\frac{n!}{(k+2)!(n-k-2)!}} = \frac{k+2}{n-k-1}\\ 8&=2n-5k\end{align*}$ This is a linear system of two equations with two unknowns, indicating that there is a unique solution. Solving by substitution or multiplying the top equation and subtracting, we find $k = 4, n = 14$ . Thus, $n+k=\boxed{018}$ .

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Mock AIME 1 2007-2008 Problems/Problem 2

From AoPSWiki

Problem

Solution

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