Question

# In the expansion of ${{\left( 1+a \right)}^{m+n}}$ prove that coefficients of ${{a}^{m}}$ and ${{a}^{n}}$ are equal.

Hint: ${{\left( 1+a \right)}^{m+n}}$ is similar to the binomial expansion of ${{\left( a+b \right)}^{n}}$. Find the expansion and substitute ${{a}^{r}}={{a}^{m}}$ and ${{a}^{r}}={{a}^{n}}$. The simplification will state that the coefficient of both ${{a}^{m}}$and ${{a}^{n}}$ is same.

Complete step-by-step Solution:

Given an expression ${{\left( 1+a \right)}^{m+n}}$. We need to prove that the expansion of ${{\left( 1+a \right)}^{m+n}}$will result in the coefficients ${{a}^{m}}$and ${{a}^{n}}$ being equal.
We know the general term of expansion of ${{\left( a+b \right)}^{n}}$, which is a binomial expansion.
It is possible to expand the polynomial ${{\left( a+b \right)}^{n}}$into a sum involving term of form $xa{{b}^{z}}$, where exponents y and z are non-negative integers and $n=y+z$, and co-efficient x of each-term is a specific positive integer.
${{\left( a+b \right)}^{n}}$ is expanded as, ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$.
i.e. if a and b are real numbers and n is a positive integer then,
${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+......+{}^{n}{{C}_{n}}{{b}^{n}}$
where, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$for $0\le r\le n$.
Therefore, general term or ${{\left( r+1 \right)}^{th}}$term in the expansion given by,
${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$
Now, for ${{\left( 1+a \right)}^{m+n}}={{\left( a+b \right)}^{n}}$.
Let’s put $n=m+n$, a = 1 and b = a.
\begin{align} & {{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}} \\ & {{T}_{r+1}}={}^{n+m}{{C}_{r}}{{1}^{\left( n+m-r \right)}}{{\left( a \right)}^{r}} \\ & {{T}_{r+1}}={}^{n+m}{{C}_{r}}{{a}^{r}}\times {{1}^{\left( n+m-r \right)}}={}^{n+m}{{C}_{r}}{{a}^{r}}\times 1 \\ \end{align}
We know ${{1}^{\left( n+m-r \right)}}$is equal to 1 i.e. 1 raised to any integer is 1.
${{T}_{r+1}}={}^{n+m}{{C}_{r}}{{a}^{r}}-(1)$
Now here we need to find the coefficients of ${{a}^{m}}$and ${{a}^{n}}$and prove that their coefficients are the same.
Finding coefficient of ${{a}^{m}}$, let us put ${{a}^{r}}={{a}^{m}}$.
$\therefore r=m$
Let us put r = m in equation (1).
${{T}_{m+1}}={}^{n+m}{{C}_{m}}{{a}^{m}}$
${}^{n+m}{{C}_{m}}$is of the form ${}^{n}{{C}_{r}}$where $\dfrac{n!}{r!\left( n-r \right)!}$.
\begin{align} & {{T}_{m+1}}=\dfrac{\left( n+m \right)!}{\left( n+m-m \right)!m!}{{a}^{m}} \\ & {{T}_{m+1}}=\dfrac{\left( n+m \right)!}{n!m!}{{a}^{m}} \\ \end{align}
Hence the coefficient of ${{a}^{m}}$is $\dfrac{\left( n+m \right)!}{n!m!}$.
Now let us find the coefficient of ${{a}^{n}}$.
Put, ${{a}^{r}}={{a}^{n}}\Rightarrow r=n$.
Put r=n in equation (1).
\begin{align} & {{T}_{r+1}}={}^{n+m}{{C}_{r}}{{a}^{r}} \\ & {{T}_{n+1}}={}^{n+m}{{C}_{n}}{{a}^{n}} \\ & {{T}_{n+1}}=\dfrac{\left( n+m \right)!}{\left( n+m-n \right)n!}\times {{a}^{n}} \\ & {{T}_{n+1}}=\dfrac{\left( n+m \right)!}{m!n!}{{a}^{n}} \\ \end{align}
Hence, the coefficient of ${{a}^{n}}$is $\dfrac{\left( n+m \right)!}{m!n!}$.
$\therefore$Coefficient of ${{a}^{m}}$= coefficient of ${{a}^{n}}=\dfrac{\left( n+m \right)!}{m!n!}$.
Hence proved.

Note:
You can consider the expansion of ${{\left( a+b \right)}^{n}}$if you know it by heart. But it is also easy to derive. Remember to put ${{a}^{r}}={{a}^{m}}$and ${{a}^{r}}={{a}^{n}}$. Then only we can prove the coefficients are the same. Also remember the expansion of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.