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

$\begin{array}{1 1}\text{They are equal} \\ \text{They are both 0} \\ a^m \text{is greater than} a^n \\ a^m \text{is lesser than} a^n\end{array} $

1 Answer

  • $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
  • Looking at the pattern of the successive terms we can say that the $(r + 1)\text{th}$ term $T_{r+1} = ^n\large C$$_r\;a^{n–r}b^r$. This is also called the general term of the expansion.
  • If $n$ is even, there will be only one middle term in the expansion of $(a+b)^n$ which is $\large\frac{n}{2} $$+1$$^{\text{th}}$ term
Assuming that $a^m$ occurs in the $(r+1)^{\text{th}}$ term of $(1+a)^{m+n}$, $T_{r+1} = ^{m+n}\large C$$_r\;1^{m+n–r}a^r$.
Comparing the indices of $a \rightarrow r = m$
Therefore, the coefficient of $a^m = \;^{m+n}\large C$$_{m} = \large\frac{(m+n)!}{m!(m+n-m)!}$$ = \large\frac{(m+n)!}{m!n!}$
Assuming that $a^n$ occurs in the $(k+1)^{\text{th}}$ term of $(1+a)^{m+n}$, $T_{k+1} = ^{m+n}\large C$$_k\;1^{m+n–k}a^l$.
Comparing the indices of $a \rightarrow k = n$,
Therefore, the coefficient of $a^n = \;^{m+n}\large C$$_{n} = \large\frac{(m+n)!}{n!(m+n-n)!}$$ = \large\frac{(m+n)!}{n!m!}$
$\Rightarrow$ we can see that the two coefficients are equal.
answered Apr 10, 2014 by balaji.thirumalai
edited Apr 10, 2014 by balaji.thirumalai

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