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

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Aicharya · Accepted Answer

We have, $ \begin{aligned} T_p & ={ }^n C_{p-1}=p \ T_{p+1} & ={ }^n C_p=q \ 	herefore \quad \frac{p}{q} & =\frac{{ }^n C_{p-1}}{{ }^n C_p} \ \Rightarrow \quad \frac{p}{q} & =\frac{p}{n-p+1} \Rightarrow p+q=n+1 \end{aligned} $

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

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