Question

Find all values of n for which (n choose (k+1)) = 3(n choose k)

Answer 1

\[ \begin{align}{n \choose k+1}&=3{n \choose k}\\ \frac{n!}{(k+1)!(n-(k+1))!}&=3\frac{n!}{k!(n-k)!} \\ \frac{\cancel{n!}}{(k+1)\cancel{k!}\cancel{(n-k-1)!}}&=3\frac{\cancel{n!}}{\cancel{k!}(n-k)\cancel{(n-k-1)!}}\\ \frac{1}{k+1}&=3\frac{1}{n-k}\\ ~ \\n-k&=3(k+1)\\ ~ \\ n&=3(k+1)+k\\ ~ \\ n&=4k+3\end{align}\]