Find the sum of the coefficients of two middle terms in the binomial expansion of $(1+x)^{2 n-1}$.
$(1+x)^{2 n-1}$
Here, $n$ is an odd number.
Therefore, the middle terms are $\left(\frac{2 n-1+1}{2}\right)^{\text {th }}$ and $\left(\frac{2 n-1+1}{2}+1\right)^{\text {th }}$, i. e.,$n^{\text {th }}$ and $(n+1)^{\text {th }}$ terms.
Now, we have
$T_{n}=T_{n-1+1}$
$={ }^{2 n-1} C_{n-1}(x)^{n-1}$
And,
$T_{n+1}=T_{n+1}$
$={ }^{2 n-1} C_{n}(x)^{n}$
$\therefore$ the coefficients of two middle terms are ${ }^{2 n-1} C_{n-1}$ and ${ }^{2 n-1} C_{n}$.
Now,
${ }^{2 \mathrm{n}-1} \mathrm{C}_{\mathrm{n}-1}+{ }^{2 \mathrm{n}-1} \mathrm{C}_{\mathrm{n}}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}$
Hence, the sum of the coefficients of two middle terms in the binomial expansion of $(1+x)^{2 n-1}$ is ${ }^{2 n} C_{n}$.