In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.

It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right)$, in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1}={ }^{n} C_{r} a^{n-t} b^{z}$.

Assuming that $a^{m}$ occurs in the $(r+1)^{\text {th }}$ term of the expansion $(1+a)^{m+n}$, we obtain

$T_{r+1}={ }^{m+n} C_{r}(1)^{m+n-r}(a)^{r}={ }^{m+n} C_{r} a^{r}$

Comparing the indices of $a$ in $a^{m}$ and in $T_{r+1}$, we obtain

$r=m$

Therefore, the coefficient of $a^{m}$ is

${ }^{m+n} C_{m}=\frac{(m+n) !}{m !(m+n-m) !}=\frac{(m+n) !}{m ! n !}$

Assuming that $a^{n}$ occurs in the $(k+1)^{t h}$ term of the expansion $(1+a)^{m+n}$, we obtain

$T_{k+1}={ }^{m+n} C_{k}(1)^{m+n-k}(a)^{k}={ }^{m+n} C_{k}(a)^{k}$

Comparing the indices of $a$ in $a^{n}$ and in $T_{k+1}$, we obtain

k = n

Therefore, the coefficient of $a^{n}$ is

${ }^{m+n} C_{n}=\frac{(m+n) !}{n !(m+n-n) !}=\frac{(m+n) !}{n ! m !} $ $\ldots(2)$

Thus, from (1) and (2), it can be observed that the coefficients of $a^{m}$ and $a^{n}$ in the expansion of $(1+a)^{m+n}$ are equal.