Question

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

• A. $4$
• B. $6$
• C. $15$
• D. $13$

Answer 1

The correct option is B $6$

Let the 3 consecutive terms in the expansion of $(1+x{)}^{n+5}$ be ${}^{n+5}{C}_{r-1}$, ${}^{n+5}{C}_r$ and ${}^{n+5}{C}_{r+1}$

Now, $\frac{{}^{n+5}{C}_{r-1}}{{}^{n+5}{C}_r}=\frac{5}{10}=\frac{1}{2}\Rightarrow \frac{(n+5)!}{(n+6-r)!(r-1)!}\frac{r!(n+5-r)!}{(n+5)!}=\frac{1}{2}\Rightarrow \frac{r}{n+6-r}=\frac{1}{2}\Rightarrow 3r=n+6...\left(i\right)$

Again, $\frac{{}^{n+5}{C}_r}{{}^{n+5}{C}_{r+1}}=\frac{10}{14}=\frac{5}{7}$

$\Rightarrow \frac{(n+5)!}{(n+5-r)!\left(r\right)!}\frac{(r+1)!(n+4-r)!}{(n+5)!}=\frac{5}{7}\Rightarrow \frac{r+1}{n+5-r}=\frac{5}{7}\Rightarrow 12r=5n+18...\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$, $n=6$