Show that the coefficient of the middle term in the expansion of $(1+x)^{2 n}$ is equal to the sum of the coefficients of two middle terms in the expansion of $(1+x)^{2 n-1}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

As $2 n$ is even so the expansion $(1+x)^{2 n}$ has only one middle term which is
$\left(\frac{2 n}{2}+1\right)^{\text {th }}$ i.e., $(n+1)^{\text {th }}$ term.

The $(n+1)^{\text {th }}$ term is $^{2 n} C_{n} x^{n}$. The coefficient of $x^{n}$ is $^{2 n} C_{n}$

Similarly, $(2 n-1)$ being odd, the other expansion has two middle terms,

$\left(\frac{2 n-1+1}{2}\right)^{ th }$ and $\left(\frac{2 n-1+1}{2}+1\right)^{ th }$ i.e., $n^{ th }$ and $(n+1)^{ th }$ terms. The coefficients of  these terms are $^{2n - 1}{C_{n - 1}}$ and $^{2n - 1}{C_n},$ respectively.

$^{2n - 1}{C_{n - 1}} + {\,^{2n - 1}}{C_n} = {\,^{2n}}{C_n}$      [ As ${^n{C_{r - 1}} + {\,^n}{C_r} = {\,^{n + 1}}{C_r}}$ ] as required.

Similar Questions

If the non zero coefficient of $(2r + 4)th$ term is greater than non zero coefficient of $(r - 2)th$ term in expansion of $(1 + x)^{18}$, then number of possible integral values of $r$ is

The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+\ldots+x^{10}$ is

  • [JEE MAIN 2020]

The coefficient of $\frac{1}{x}$ in the expansion of  ${\left( {1 + x} \right)^n}{\left( {1 + \frac{1}{x}} \right)^n}$ is :-

The term independent of $x$ in ${\left( {\sqrt x - \frac{2}{x}} \right)^{18}}$ is

If coefficient of ${(2r + 3)^{th}}$ and ${(r - 1)^{th}}$ terms in the expansion of ${(1 + x)^{15}}$ are equal, then value of r is