Question

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

• A. $792$
• B. $252$
• C. $462$
• D. $330$

Answer 1

The correct option is C

$462$

Finding largest coefficient in the given expansion:

Step 1: Finding the general terms

The ratio of three consecutive term is $5:10:14$

Let three consecutive terms are ${\mathrm{T}}_{\mathrm{r},}{\mathrm{T}}_{\mathrm{r}+1,}{\mathrm{T}}_{\mathrm{r}+2}$

So, $\frac{{\mathrm{T}}_{\mathrm{r}}}{{\mathrm{T}}_{\mathrm{r}+1}}=\frac{5}{10}\mathrm{and}\frac{{\mathrm{T}}_{\mathrm{r}+1}}{{\mathrm{T}}_{\mathrm{r}+2}}=\frac{10}{14}$

Here, the coefficient ${\mathrm{T}}_{\mathrm{r},}{\mathrm{T}}_{\mathrm{r}+1,}{\mathrm{T}}_{\mathrm{r}+2}$can be written as,

$$\begin{gathered} \mathrm{Tr}=\mathrm{n}+5_{\mathrm{cr}-1,}{\mathrm{T}}_{\mathrm{r}+1}=\mathrm{n}+5_{\mathrm{cr}},{\mathrm{T}}_{\mathrm{r}+2}=\mathrm{n}+5_{\mathrm{cr}+1} \\ \therefore \frac{{\mathrm{T}}_{\mathrm{r}+1}}{{\mathrm{T}}_{\mathrm{r}}}=2\mathrm{and}\frac{{\mathrm{T}}_{\mathrm{r}+1}}{{\mathrm{T}}_{\mathrm{r}+2}}=\frac{5}{7} \\ \frac{\mathrm{n}+5_{\mathrm{cr}}}{\mathrm{n}+5_{\mathrm{cr}-1}}=2\mathrm{and}\frac{\mathrm{n}+5_{\mathrm{cr}}}{\mathrm{n}+5_{\mathrm{cr}+1}}=\frac{5}{7} \end{gathered}$$

Step 2: Solving $\frac{\mathrm{n}+5_{\mathrm{cr}}}{\mathrm{n}+5_{\mathrm{cr}-1}}=2$

${\mathrm{n}}_{\mathrm{cr}}$ can be written as,

${\mathrm{n}}_{\mathrm{cr}}=\frac{\mathrm{n}!}{\mathrm{r}(\mathrm{n}-\mathrm{r})!}$

Using this in above equation,

$\begin{array}{rcl}\frac{{\displaystyle \frac{(\mathrm{n}+5)!}{\mathrm{r}!(\mathrm{n}+5-\mathrm{r})!}}}{{\displaystyle \frac{(\mathrm{n}+5)!}{(\mathrm{r}-1)!(\mathrm{n}+5-\mathrm{r}+1)!}}}& =& 2\\ & \Rightarrow & \frac{(\mathrm{r}-1)!(\mathrm{n}+5-\mathrm{r}+1)!}{\mathrm{r}!(\mathrm{n}+5-\mathrm{r})!}=2\\ & \Rightarrow & \frac{(\mathrm{n}+5)-\mathrm{r}+1}{\mathrm{r}}=2\\ & \Rightarrow & \mathrm{n}-\mathrm{r}+6=2\mathrm{r}\\ \therefore \mathrm{n}-3\mathrm{r}+6& =& 0\to \left(1\right)\end{array}$

Step 3: Solving $\frac{\mathrm{n}+5_{\mathrm{cr}}}{\mathrm{n}+5_{\mathrm{cr}+1}}=\frac{5}{7}$

$$\begin{gathered} \begin{array}{rcl}\frac{{\displaystyle \frac{(\mathrm{n}+5)!}{(\mathrm{r}+1)!(\mathrm{n}+5-\mathrm{r}-1)!}}}{{\displaystyle \frac{(\mathrm{n}+5)!}{\mathrm{r}!(\mathrm{n}+5-\mathrm{r})!}}}& =& \frac{7}{5}\\ & \Rightarrow & \frac{(\mathrm{n}+5)-(\mathrm{r}+1)+1}{\mathrm{r}+1}=\frac{7}{5}\\ & \Rightarrow & \frac{\mathrm{n}-\mathrm{r}+5}{\mathrm{r}+1}=\frac{7}{5}\\ & \Rightarrow & 5\mathrm{n}-5\mathrm{r}+25=7\mathrm{r}+7\end{array} \\ \therefore 5n-12r+18=0\to \left(2\right) \end{gathered}$$

Step 4: Solving equation (1) and equation (2) to find $r$

Now, multiply equation $\left(1\right)$ by $5,$ and subtracting with equation $\left(2\right),$

$$\begin{gathered} \begin{array}{rcl}& & \frac{5\mathrm{n}-15\mathrm{r}+30=0 \\ 5\mathrm{n}-12\mathrm{r}+18=0}{-3\mathrm{r}+12=0}\\ & \Rightarrow & -3\mathrm{r}+12=0\\ & \Rightarrow & -3\mathrm{r}=-12\\ \therefore \mathrm{r}& =& 4\end{array} \end{gathered}$$

Step 5: Finding the value of $‘n’$ by substituting $‘r’$ value

Substitute $‘r’$ value in equation $\left(1\right)$

$\begin{array}{rcl}5n-15r+30& =& 0\\ 5n-15\left(4\right)+30& =& 0\\ 5n-60+30& =& 0\\ 5n-30& =& 0\\ 5n& =& 30\\ n& =& 6\\ & & \end{array}$

Step 6: Finding largest coefficient in the given expansion

Given expansion ${(1+\mathrm{x})}^{\mathrm{n}+5}={(1+\mathrm{x})}^{6+5}{(1+\mathrm{x})}^{11}={11}_{\mathrm{c}6}$

this can be written as,

$$\begin{gathered} {11}_{\mathrm{c}6}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!} \\ {11}_{\mathrm{c}6}=\frac{11!}{6!(11-6)!} \\ {11}_{\mathrm{c}6}=\frac{11!}{6!\left(5\right)!} \\ {11}_{\mathrm{c}6}=462 \end{gathered}$$

Therefore, the largest coefficient in the expansion is 462.

Hence, the correct answer is option(c).