If some three consecutive in the binomial expansion of ${{\left( x+1 \right)}^{n}}$ is powers of x are in the ratio 2: 15: 70, then the average of these three coefficients is:
a. 964
b. 625
c. 227
d. 232

Hint: We will use the Newton’s Binomial expansion for this question which states that,
${{\left( x+y \right)}^{n}}=\sum\limits_{j=0}^{n}{\left( \begin{align}
  & n \\
 & k \\
\end{align} \right){{x}^{j}}{{y}^{n-j}}}$ and after that we take the ratio of the terms and equate it to the given ratio’s to find the values of n . Using the values of n, we will find the values of the terms and find their average.

Complete step-by-step solution
By Newton’s Binomial Theorem, we have,
${{\left( x+y \right)}^{n}}=\sum\limits_{j=0}^{n}{\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right){{x}^{j}}{{y}^{n-j}}}$
Let us assume that the three consecutive terms are the ${{\left( k-1 \right)}^{th}},{{k}^{th}},{{\left( k+1 \right)}^{th}}$ terms. The coefficients of these terms are given by,
$\begin{align}
  & \Rightarrow {{t}_{1}}=\left( \begin{matrix}
   n \\
   k-1 \\
\end{matrix} \right) \\
 & \Rightarrow {{t}_{2}}=\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right) \\
 & \Rightarrow {{t}_{3}}=\left( \begin{matrix}
   n \\
   k+1 \\
\end{matrix} \right) \\
\end{align}$
According to the question \[{{t}_{1}}:{{t}_{2}}:{{t}_{3}}=2:15:70\].
$\begin{align}
  & \Rightarrow \dfrac{{{t}_{1}}}{{{t}_{2}}}=\dfrac{2}{15} \\
 & \Rightarrow \dfrac{{{t}_{2}}}{{{t}_{3}}}=\dfrac{15}{70}=\dfrac{3}{14} \\
 & \Rightarrow \dfrac{{{t}_{1}}}{{{t}_{3}}}=\dfrac{2}{70}=\dfrac{1}{35} \\
\end{align}$
Using the first equation, we get that,
$\Rightarrow \dfrac{\left( \begin{matrix}
   n \\
   k-1 \\
\end{matrix} \right)}{\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right)}=\dfrac{2}{15}$
Simplifying we get,
$\begin{align}
  & \Rightarrow \dfrac{\dfrac{n!}{\left( k-1 \right)!\left( n-k+1 \right)!}}{\dfrac{n!}{\left( k \right)!\left( n-k \right)!}}=\dfrac{2}{15} \\
 & \Rightarrow \dfrac{\dfrac{n!}{\left( k-1 \right)!\left( n-k+1 \right)\left( n-k \right)!}}{\dfrac{n!}{k\left( k-1 \right)!\left( n-k \right)!}}=\dfrac{2}{15} \\
\end{align}$
On cancelling the common terms in the LHS, we will get,
$\Rightarrow \dfrac{k}{\left( n-k+1 \right)}=\dfrac{2}{15}$
On cross multiplying, we get,
$\begin{align}
  & \Rightarrow 15k=2\left( n-k+1 \right) \\
 & \Rightarrow 15k=2n-2k+2 \\
 & \Rightarrow 2n=17k-2.........(a) \\
\end{align}$
Similarly from the second equation, we have that,
$\Rightarrow \dfrac{\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right)}{\left( \begin{matrix}
   n \\
   k+1 \\
\end{matrix} \right)}=\dfrac{3}{14}$
Simplifying we get,
$\begin{align}
  & \Rightarrow \dfrac{\dfrac{n!}{\left( k \right)!\left( n-k \right)!}}{\dfrac{n!}{\left( k+1 \right)!\left( n-k-1 \right)!}}=\dfrac{3}{14} \\
 & \Rightarrow \dfrac{\dfrac{n!}{k!\left( n-k \right)\left( n-k-1 \right)!}}{\dfrac{n!}{\left( k+1 \right)k!\left( n-k-1 \right)!}}=\dfrac{3}{14} \\
\end{align}$
On cancelling the like terms in the LHS, we get,
$\Rightarrow \dfrac{k+1}{\left( n-k \right)}=\dfrac{3}{14}$
On cross multiplying, we get,
$\begin{align}
  & \Rightarrow 14\left( k+1 \right)=3\left( n-k \right) \\
 & \Rightarrow 14k+14=3n-3k \\
 & \Rightarrow 3n=17k+14.........(b) \\
\end{align}$
From these two equations, we will get the following relations between n and k.
On subtracting (a) from (b), we get,
$\begin{align}
  & \Rightarrow 3n-2n=17k+14-17k+2 \\
 & \Rightarrow n=16 \\
\end{align}$
On substituting n = 16 in equation (b), we get,
$\begin{align}
  & \Rightarrow 3\times 16=17k+14 \\
 & \Rightarrow 48=17k+14 \\
 & \Rightarrow 17k=34 \\
 & \Rightarrow k=\dfrac{34}{17}=2 \\
\end{align}$
Hence, we get n = 16 and k = 2.
We need to find the average of the coefficients of these terms. The required average A is given by,
$\Rightarrow A=\dfrac{1}{3}\left[ \left( \begin{matrix}
   n \\
   k-1 \\
\end{matrix} \right)+\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right)+\left( \begin{matrix}
   n \\
   k+1 \\
\end{matrix} \right) \right]$
Substituting the value of n and k in this equation, we get that,
\[\begin{align}
  & \Rightarrow A=\dfrac{1}{3}\left[ \left( \begin{matrix}
   16 \\
   1 \\
\end{matrix} \right)+\left( \begin{matrix}
   16 \\
   2 \\
\end{matrix} \right)+\left( \begin{matrix}
   16 \\
   3 \\
\end{matrix} \right) \right] \\
 & \Rightarrow \dfrac{1}{3}\left[ \dfrac{16!}{1!\left( 16-1 \right)!}+\dfrac{16!}{2!\left( 16-2 \right)!}+\dfrac{16!}{3!\left( 16-3 \right)!} \right] \\
 & \Rightarrow \dfrac{1}{3}\left[ \dfrac{16\times \left( 15 \right)!}{\left( 15 \right)!}+\dfrac{16\times 15\times \left( 14 \right)!}{2\times \left( 14 \right)!}+\dfrac{16\times 15\times 14\times \left( 13 \right)!}{3\times 2\times \left( 13 \right)!} \right] \\
 & \Rightarrow \dfrac{1}{3}\left[ 16+120+560 \right] \\
 & \Rightarrow \dfrac{1}{3}\left( 696 \right) \\
 & \Rightarrow 232 \\
\end{align}\]
Hence, we get the answer as 232. Therefore, option (d) is the correct answer.

Note: The key part in this question is to use Newton’s Binomial Theorem to find the general term in the expansion of ${{\left( x+1 \right)}^{n}}$ and then compare the ratio of the consecutive terms to the given ratio to find the terms and then find the average. It is important to remember that the kth term of the binomial expansion is obtained by substituting j = k − 1 in Newton’s Binomial Theorem. So if you substitute j = k, you will end up with the wrong answer.