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

$(1+x)^{n+5} = \sum_{r} \binom{n+5}{r}\, x^r$ $\binom{n+5}{k} : \binom{n+5}{k+1} : \binom{n+5}{k+2} = 5 : 10 : 14$ $\text{To find: } n$ $a = \binom{N}{k},\quad b = \binom{N}{k+1},\quad c = \binom{N}{k+2}\quad (N = n+5)$ $a : b : c = 5 : 10 : 14$ $\dfrac{b}{a} = \dfrac{\binom{N}{k+1}}{\binom{N}{k}} = \dfrac{N-k}{k+1} = \dfrac{10}{5} = 2$ $\Rightarrow N - k = 2(k+1) \Rightarrow N = 3k + 2$ $\dfrac{c}{b} = \dfrac{\binom{N}{k+2}}{\binom{N}{k+1}} = \dfrac{N-k-1}{k+2} = \dfrac{14}{10} = \dfrac{7}{5}$ $5(N-k-1) = 7(k+2) \Rightarrow 5N = 12k + 19$ $5(3k+2) = 12k + 19$ $15k + 10 = 12k + 19 \Rightarrow 3k = 9 \Rightarrow k = 3$ $N = 3(3) + 2 = 11$ $N = n+5 \Rightarrow n = 11 - 5 = 6$ $\binom{11}{3} : \binom{11}{4} : \binom{11}{5} = 165 : 330 : 462 = 5 : 10 : 14\ \checkmark$ $n = 6$