Question

If in the expansion of (1 + x)ⁿ, the coefficients of pth and qth terms are equal, prove that p + q = n + 2, where $p\ne q$.

Answer 1

$$\begin{gathered} \mathrm{Coefficients}\mathrm{of}\mathrm{the}p\mathrm{th}\mathrm{and}q\mathrm{th}\mathrm{terms}\mathrm{are}{}^{n}C_{p-1}\mathrm{and}{}^{n}C_{q-1}\mathrm{respectively}. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ {}^{n}C_{p-1}={}^{n}C_{q-1} \\ \Rightarrow p-1=q-1\mathrm{or},p-1+q-1=n[\because {}^{n}C_r{=}^n{C}_s\Rightarrow r=s\mathrm{or},r+s=n] \\ \Rightarrow p=q\mathrm{or},p+q=n+2 \end{gathered}$$

If $p\ne q$, then $p+q=n+2$
Hence proved