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

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**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 \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].

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