Question

If the coefficient of $r^{th}$, \[(r+1)^{th}\], and \[(r+2)^{th}\] term in the binomial expansion of ${{(1+y)}^{m}}$ are in the arithmetic progression, then $m$ and $r$ satisfy the equation
A. ${{m}^{2}}-m(4r-1)+4{{r}^{2}}-2=0$
B. ${{m}^{2}}-m(4r+1)+4{{r}^{2}}+2=0$
C. ${{m}^{2}}-m(4r+1)+4{{r}^{2}}-2=0$
D. ${{m}^{2}}-m(4r-1)+4{{r}^{2}}+2=0$

Answer 1

Hint: In this question, we are to find the equation formed by the variables $m$ and $r$. Here these are the variables from a binomial expression ${{(1+y)}^{m}}$. So, by using the binomial theorem we can find the required term to form an arithmetic series as given in the question. Then, by the arithmetic mean we can frame the required equation.

Formula used: If the series is an Arithmetic series, then the sum of the $n$ terms of the arithmetic series is calculated by
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
Where the common difference $d={{a}_{n}}-{{a}_{n-1}}$
The arithmetic mean of the terms $a,b,c$ which are in the arithmetic progression is
$2b=a+c$
According to the binomial theory, the expansion of a binomial expression is
${{(x+a)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+...+{}^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}+...+{}^{n}{{C}_{n}}{{a}^{n}}$
We can also write it as
${{(x+a)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}}$
Here the expansion contains $(n+1)$ terms and the $(r+1)^{th}$ term is called the general term of the expansion. I.e.,
${{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}$
Here ${}^{n}{{C}_{r}}$ represents the coefficient of the term and it is calculated by the combinations and it is formulated as
${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$
Some of the important combination formulae are:
\[\begin{align}
  & n!=n(n-1)(n-2)...(n-r+1)=n\cdot (n-1)! \\
 & {}^{n}{{C}_{n}}=1 \\
 & {}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}} \\
 & {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{(n+1)}{{C}_{r}} \\
 & {}^{n}{{C}_{r}}={}^{n}{{C}_{s}}\Leftrightarrow r=s;r+s=n \\
\end{align}\]

Complete step by step solution: The given binomial expression is ${{(1+y)}^{m}}$
Where $a=1;x=y;n=m$
According to the binomial theorem, the given expression is expanded as
${{(x+a)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+...+{}^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}+...+{}^{n}{{C}_{n}}{{a}^{n}}$
${{(1+y)}^{m}}={}^{m}{{C}_{0}}{{y}^{m}}+{}^{m}{{C}_{1}}{{y}^{m-1}}(1)+{}^{m}{{C}_{2}}{{y}^{m-2}}{{(1)}^{2}}+...+{}^{m}{{C}_{r}}{{y}^{m-r}}{{(1)}^{r}}+...+{}^{m}{{C}_{m}}{{(1)}^{m}}\text{ }...(1)$
From (1), we can write the required terms as:
The $r^{th}$ term, \[(r+1)^{th}\], and \[(r+2)^{th}\] of the above expansion are obtained by the formula,
${{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}$
On substituting the respective values into the above formula, we get
$\begin{align}
  & {{T}_{r}}={{T}_{(r-1)+1}}={}^{m}{{C}_{(r-1)}}{{y}^{m-(r-1)}}{{(1)}^{(r-1)}} \\
 & {{T}_{r+1}}={}^{m}{{C}_{r}}{{y}^{m-r}}{{(1)}^{r}} \\
 & {{T}_{r+2}}={{T}_{(r+1)+1}}={}^{m}{{C}_{(r+1)}}{{y}^{m-(r+1)}}{{(1)}^{r+1}} \\
\end{align}$
From the above terms, the coefficients are
The coefficient of $r^{th}$ term is ${}^{m}{{C}_{r-1}}$, the coefficient of \[(r+1)^{th}\] term is ${}^{m}{{C}_{r}}$, and the coefficient of \[(r+2)^{th}\] term is ${}^{m}{{C}_{r+1}}$.
Since it is given that, these coefficient ${}^{m}{{C}_{r-1}}$, ${}^{m}{{C}_{r}}$, ${}^{m}{{C}_{r+1}}$ are in arithmetic progression, we relate them with their arithmetic mean as
$2b=a+c$ i.e.,
\[2{}^{m}{{C}_{r}}={}^{m}{{C}_{r-1}}+{}^{m}{{C}_{r+1}}\text{ }...(2)\]
Thus, on simplifying (2), we get
\[\begin{align}
  & 2{}^{m}{{C}_{r}}={}^{m}{{C}_{r-1}}+{}^{m}{{C}_{r+1}} \\
 & \Rightarrow \dfrac{2{}^{m}{{C}_{r}}}{{}^{m}{{C}_{r}}}=\dfrac{{}^{m}{{C}_{r-1}}}{{}^{m}{{C}_{r}}}+\dfrac{{}^{m}{{C}_{r+1}}}{{}^{m}{{C}_{r}}} \\
 & \Rightarrow 2=\dfrac{{}^{m}{{C}_{r-1}}}{{}^{m}{{C}_{r}}}+\dfrac{{}^{m}{{C}_{r+1}}}{{}^{m}{{C}_{r}}}\text{ }...(3) \\
\end{align}\]
Here, we have
\[\begin{align}
  & {}^{m}{{C}_{r-1}}=\dfrac{m!}{(r-1)!(m-r+1)!} \\
 & {}^{m}{{C}_{r}}=\dfrac{m!}{r!(m-r)!} \\
 & {}^{m}{{C}_{r+1}}=\dfrac{m!}{(r+1)!(m-r-1)!} \\
\end{align}\]
Then, we can calculate
$\begin{align}
  & \dfrac{{}^{m}{{C}_{r-1}}}{{}^{m}{{C}_{r}}}=\dfrac{m!}{(r-1)!(m-r+1)!}\times \dfrac{r!(m-r)!}{m!}=\dfrac{r(r-1)!(m-r)!}{(r-1)!(m-r+1)(m-r)!}=\dfrac{r}{m-r+1} \\
 & \dfrac{{}^{m}{{C}_{r+1}}}{{}^{m}{{C}_{r}}}=\dfrac{m!}{(r+1)!(m-r-1)!}\times \dfrac{r!(m-r)!}{m!}=\dfrac{r!(m-r)(m-r-1)!}{(r+1)r!(m-r-1)!}=\dfrac{m-r}{r+1} \\
\end{align}$
On substituting these values into (3), we get
\[\begin{align}
  & 2=\dfrac{{}^{m}{{C}_{r-1}}}{{}^{m}{{C}_{r}}}+\dfrac{{}^{m}{{C}_{r+1}}}{{}^{m}{{C}_{r}}} \\
 & \Rightarrow 2=\dfrac{r}{m-r+1}+\dfrac{m-r}{r+1} \\
 & \Rightarrow 2(m-r+1)(r+1)=r(r+1)+(m-r)(m-r+1) \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 2(mr+m-{{r}^{2}}-r+r+1)={{r}^{2}}+r+{{m}^{2}}-mr+m-mr+{{r}^{2}}-r \\
 & \Rightarrow 2mr+2m-2{{r}^{2}}+2=2{{r}^{2}}+{{m}^{2}}-2mr+m \\
 & \Rightarrow {{m}^{2}}-4mr-m+4{{r}^{2}}-2=0 \\
 & \Rightarrow {{m}^{2}}-m(4r+1)+4{{r}^{2}}-2=0 \\
\end{align}\]
Therefore, the required equation is ${{m}^{2}}-m(4r+1)+4{{r}^{2}}-2=0$.

Thus, Option (C) is correct.

Note: Here, we need to remember that the given terms are the coefficients of the given binomial expression. So, we need to apply the binomial theorem in order to get the required terms and their coefficients. So, by simplifying them with the respective formulae, we can get the required equation.