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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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Verified Answer
The correct answer is:
$n+1$
We have,
$$
\begin{aligned}
T_p & ={ }^n C_{p-1}=p \\
T_{p+1} & ={ }^n C_p=q \\
\therefore \quad \frac{p}{q} & =\frac{{ }^n C_{p-1}}{{ }^n C_p} \\
\Rightarrow \quad \frac{p}{q} & =\frac{p}{n-p+1} \Rightarrow p+q=n+1
\end{aligned}
$$
$$
\begin{aligned}
T_p & ={ }^n C_{p-1}=p \\
T_{p+1} & ={ }^n C_p=q \\
\therefore \quad \frac{p}{q} & =\frac{{ }^n C_{p-1}}{{ }^n C_p} \\
\Rightarrow \quad \frac{p}{q} & =\frac{p}{n-p+1} \Rightarrow p+q=n+1
\end{aligned}
$$
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