Question

For a positive integer \(n\),
\(\left(1+\frac{1}{x}\right)^{n}\) is expanded
in increasing powers of \(x\). If three consecutive
coefficients in this expansion are in the ratio,
\(2 : 5 : 12\), then \(n\) is equal to _______.

Hint:

The coefficients are binomial coefficients \(\binom{n}{k}\); use \(\binom{n}{k+1}:\binom{n}{k}:\binom{n}{k-1}=2:5:12\).

Answer 1

Correct Answer: 118

The general term is \(\binom{n}{k}x^{-k}\), so in increasing powers of \(x\) the coefficients appear as \(\binom{n}{k+1},\binom{n}{k},\binom{n}{k-1}\) in the ratio \(2:5:12\).
\(\frac{\binom{n}{k}}{\binom{n}{k+1}}=\frac{k+1}{n-k}=\frac52\Rightarrow 2k+2=5n-5k\Rightarrow 7k=5n-2\).
\(\frac{\binom{n}{k-1}}{\binom{n}{k}}=\frac{k}{n-k+1}=\frac{12}{5}\Rightarrow 5k=12n-12k+12\Rightarrow 17k=12n+12\).
Substitute \(k=\frac{5n-2}{7}\): \(17(5n-2)=7(12n+12)\Rightarrow 85n-34=84n+84\Rightarrow n=118\) (and \(k=84\)).
Answer: 118