The coefficients of three consecutive terms of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then $n =$ ___

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Solution

Let the coefficients of three consecutive terms of $(1 + x)^{n + 5}$ be
$^{n + 5} C_{r - 1},^{n + 5} C_{r},^{n + 5} C_{r + 1}, then we have^{n + 5} C_{r - 1} :^{n + 5} C_{r} :^{n + 5} C_{r + 1} = 5 : 10 : 14 \frac{^{n + 5} C_{r - 1}}{^{n + 5} C_{r}} = \frac{5}{10} \Rightarrow \frac{r}{n + 6 - r} = \frac{1}{2} or n - 3 r + 6 = 0 (1) Also \frac{^{n + 1} C_{r}}{^{n + 5} C_{r + 1}} = \frac{10}{14} \Rightarrow \frac{r + 1}{n - r + 5} = \frac{5}{7} or 5 n - 12 r + 18 = 0 (2) Solving (1) and (2) we get n = 6.$

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