Question

Find the sum of the coefficients of two middle terms in the binomial expansion of ${\left(1+x\right)}^{2n-1}$.

Answer 1

$$\begin{gathered} {\left(1+x\right)}^{2n-1} \\ \mathrm{Here},n\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{number}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{middle}\mathrm{terms}\mathrm{are}{\left(\frac{2n-1+1}{2}\right)}^{\mathrm{th}}\mathrm{and}{\left(\frac{2n-1+1}{2}+1\right)}^{\mathrm{th}},\mathrm{i}.\mathrm{e}.,n^{\mathrm{th}}\mathrm{and}(n+1{)}^{\mathrm{th}}\mathrm{terms}. \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {T}_n=T_{n-1+1} \\ ={}^{2n-1}C_{n-1}{\left(x\right)}^{n-1} \\ \mathrm{And}, \\ {T}_{n+1}=T_{n+1} \\ ={}^{2n-1}C_n{\left(x\right)}^n \\ \therefore \mathrm{the}\mathrm{coefficients}\mathrm{of}\mathrm{two}\mathrm{middle}\mathrm{terms}\mathrm{are}{}^{2n-1}C_{n-1}\mathrm{and}{}^{2n-1}C_n. \\ \mathrm{Now},{ \\ }^{2\mathrm{n}-1}{\mathrm{C}}_{\mathrm{n}-1}{+}^{2\mathrm{n}-1}{\mathrm{C}}_{\mathrm{n}}{=}^{2\mathrm{n}}{\mathrm{C}}_{\mathrm{n}} \end{gathered}$$

Hence, the sum of the coefficients of two middle terms in the binomial expansion of ${\left(1+x\right)}^{2n-1}$ is ${}^{2n}{C}_n$.