If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)^{n + 5} are in the ratio 5 : 10 : 14, then the
Iargest coefficient in this expansion is

Solution:

 Let the three consecutive terms in the binomial expansion of $(1+\mathrm{x}{)}^{\mathrm{n}+5}$ are

${ }^{\mathrm{n}+5}{\mathrm{C}}_{\mathrm{r}-1}{,}^{\mathrm{n}+5}{\mathrm{C}}_{\mathrm{r}}{\text{ and }}^{\mathrm{n}+5}{\mathrm{C}}_{\mathrm{r}+1}$ 

Now, according to the given information ${ }^{\mathrm{n}+5}{\mathrm{C}}_{\mathrm{r}-1}{:}^{\mathrm{n}+5}{\mathrm{C}}_{\mathrm{r}}{:}^{\mathrm{n}+5}{\mathrm{C}}_{\mathrm{r}+1}=5:10:14$

$\begin{array}{r}\Rightarrow \frac{(\mathrm{n}+5)!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r}+6)!}:\frac{(\mathrm{n}+5)!}{\mathrm{r}!(\mathrm{n}-\mathrm{r}+5)!}:\frac{(\mathrm{n}+5)!}{(\mathrm{r}+1)!(\mathrm{n}-\mathrm{r}+4)!}=5:10:14\\ \end{array}$

$\begin{array}{l}\begin{array}{rl} & \Rightarrow \frac{1}{(\mathrm{n}-\mathrm{r}+6)(\mathrm{n}-\mathrm{r}+5)}:\frac{1}{\mathrm{r}(\mathrm{n}-\mathrm{r}+5)}:\frac{1}{(\mathrm{r}+1)\mathrm{r}}=5:10:14\\ & \text{ So, }\frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+6}=\frac{5}{10}\\ & \Rightarrow 2\mathrm{r}=\mathrm{n}-\mathrm{r}+6\Rightarrow \mathrm{n}+6=3\mathrm{r}\\ & \text{ and }\frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}+5}=\frac{5}{7}\end{array}\\ \begin{array}{lrll}\Rightarrow & & 7\mathrm{r}+7& =5\mathrm{n}-5\mathrm{r}+25\\ \Rightarrow & & 5\mathrm{n}+18& =12\mathrm{r}\end{array}\end{array}$

From Eqs. (i) and (ii), we have n =6.
So, the largest coefficient in the expansion is same as 

the greatest binomial coefficient 

$\begin{array}{l}{=}^{11}{\mathrm{C}}_5{\text{ or }}^{11}{\mathrm{C}}_6\\ =\frac{11!}{5!6!}=\frac{11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2}=462\end{array}$