In the expansion of $(1+x)^n$, the coefficients of $p$ th and $(p+1)$ th terms are respectively $p$ and $q$ then $p+q=$

In the expansion of $(1+x)^n$, the general term i.e. $(r+1)$ th tem is $$ T_{Y+1}={ }^n C_Y(1)^{n-y} X^Y={ }^n C_Y X^Y $$ $\therefore$ Coefficient of $(r+1)$ th term is ${ }^n C_Y$ Similarly, coefficient of $p$ th term $={ }^n C_{p-1}$ $$ \begin{aligned} \therefore \quad p & ={ }^n C_{p-1} \text { (given) } \\ p & =\frac{n !}{(p-1) !(n-P+1) !} \end{aligned} $$ and coefficient of $\left(p+1\right.$ th term $={ }^n C_p$ $\therefore \quad q={ }^n C_p$ (given) On dividing Eq. (i) by Eq. (ii), we get $$ \begin{aligned} & \frac{p}{q}=\frac{\frac{n !}{(p-1) !(n-p+1) !}}{{ }^n C_p}=\frac{\frac{n !}{(p-1) !(n-p+1) !}}{\frac{n !}{(p) !(n-p) !}} \\ & =\frac{(p) !(n-p) !}{(n-p+1) !(p-1) !}=\frac{p(p-1) !(n-p) !}{(n-p+1)(n-p) !(p-1) !} \end{aligned} $$ $$ \begin{array}{lc} \Rightarrow & \frac{p}{q}=\frac{p}{n-p+1} \\ \Rightarrow & \frac{1}{q}=\frac{1}{n-p+1} \\ \Rightarrow & n-p+1=q \\ \Rightarrow & p+q=n+1 \end{array} $$