$(1+a)^{m+n}$ के प्रसार में सिद्ध कीजिए कि $a^{m}$ तथा $a^{n}$ के गुणांक बराबर हैं |

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 $T_{r+1}=^{n} C_{r} a^{n-r} b^{r}$

Assuming that $a^{m}$ occurs in the $(r+1)^{th}$ 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 $a^{m}$ in $T_{r+1},$

We obtain $r = m$

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

${\,^{m + n}}{C_m} = \frac{{(m + n)!}}{{m!(m + n - m)!}} = \frac{{(m + n)!}}{{m!n!}}$       ...........$(1)$

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

${\,^{m + n}}{C_n} = \frac{{(m + n)!}}{{n!(m + n - n)!}} = \frac{{(m + n)!}}{{n!m!}}$         ............$(2)$

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