Question

If the coefficients of three consecutive terms in the expansion of ${\left(1+a\right)}^n$ are in the ratio $1:7:42$ then $n$ is divisible by

Answer 1

Step 1: Use the general term of binomial expansion to find coefficient of three consecutive terms

The binomial expansion formula is

${\left(a+b\right)}^n=\sum _{i=0}^n{}^{n}C_i{\left(a\right)}^{n-i}{\left(b\right)}^i$

Let the three consecutive term be

${\left(r-1\right)}^{th}\mathrm{term},{\left(\mathrm{r}\right)}^{\mathrm{th}}\mathrm{term}\mathrm{and}{\left(\mathrm{r}+1\right)}^{\mathrm{th}}\mathrm{term}$

The coefficient of ${\left(r-1\right)}^{th}\mathrm{term}$ is

${}^n{C}_{\left(r-1\right)-1}={}^n{C}_{r-2}$

The coefficient of ${\left(r\right)}^{th}\mathrm{term}$ is

${}^n{C}_{\left(r\right)-1}={}^n{C}_{r-1}$

The coefficient of ${\left(r+1\right)}^{th}\mathrm{term}$ is

${}^n{C}_{\left(r+1\right)-1}={}^n{C}_r$

Step 2: Form equation using given ratio

The given ratio of the coefficient is $1:7:42$

Taking the ${\left(r-1\right)}^{th}\mathrm{term}$ and ${\left(r\right)}^{th}\mathrm{term}$ we have

$$\begin{gathered} \frac{{}^n{C}_{r-2}}{{}^n{C}_{r-1}}=\frac{1}{7} \\ \Rightarrow \frac{{\displaystyle \frac{n!}{\left(r-2\right)!\left(n-r+2\right)!}}}{\frac{n!}{\left(r-1\right)!\left(n-r+1\right)!}}=\frac{1}{7} \\ \Rightarrow \frac{\left(r-1\right)!\left(n-r+1\right)!}{\left(r-2\right)!\left(n-r+2\right)!}=\frac{1}{7} \\ \Rightarrow \frac{\left(r-1\right)}{\left(n-r+2\right)}=\frac{1}{7} \\ \Rightarrow 7r-7=n-r+2 \\ \Rightarrow n-8r+9=0...................\left(1\right) \end{gathered}$$

Taking the ${\left(r\right)}^{th}\mathrm{term}$ and ${\left(r+1\right)}^{th}\mathrm{term}$ we have

$$\begin{gathered} \frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{7}{42} \\ \Rightarrow \frac{{\displaystyle \frac{n!}{\left(r-1\right)!\left(n-r+1\right)!}}}{\frac{n!}{\left(r\right)!\left(n-r\right)!}}=\frac{1}{6} \\ \Rightarrow \frac{\left(r\right)!\left(n-r\right)!}{\left(r-1\right)!\left(n-r+1\right)!}=\frac{1}{6} \\ \Rightarrow \frac{\left(r\right)}{\left(n-r+1\right)}=\frac{1}{6} \\ \Rightarrow 6r=n-r+1 \\ \Rightarrow n-7r+1=0...................\left(2\right) \end{gathered}$$

Step 3: Solve the obtained equations

Subtracting $\left(1\right)$ and $\left(2\right)$ we have

$$\begin{gathered} r-8=0 \\ \Rightarrow r=8 \end{gathered}$$

Putting the value of $r$ in equation $\left(1\right)$ we have

$\begin{array}{rcl}n-8\times 8+9& =& 0\\ & \Rightarrow & n=55\end{array}$

Hence the value of $n$is $55$