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 $(1+x)^{2 n-1}$
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 $(1+x)^{2 n-1}$
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.
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- [JEE MAIN 2020]