In the expansion of (1 + x)^n,the coefficient | vedclass

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(B) The $p^{th}$ term in the expansion of $(1 + x)^n$ is $T_p = {}^nC_{p-1} x^{p-1}$. The coefficient is $p = {}^nC_{p-1}$.The $(p + 1)^{th}$ term is

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$n + 1$

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(B) The $p^{th}$ term in the expansion of $(1 + x)^n$ is $T_p = {}^nC_{p-1} x^{p-1}$. The coefficient is $p = {}^nC_{p-1}$.The $(p + 1)^{th}$ term is $T_{p+1} = {}^nC_p x^p$. The coefficient is $q = {}^nC_p$.We know that $\frac{q}{p} = \frac{{}^nC_p}{{}^nC_{p-1}} = \frac{n - p + 1}{p}$.Thus,$pq = p(n - p + 1)$ is not the correct path; rather,we use the property of binomial coefficients:$\frac{q}{p} = \frac{n - p + 1}{p} \implies qp = p(n - p + 1)$.Actually,the question implies the coefficients are $p$ and $q$ as variables. Given $p = {}^nC_{p-1}$ and $q = {}^nC_p$,the relation $\frac{q}{p} = \frac{n - p + 1}{p}$ leads to $q = \frac{n - p + 1}{p} \cdot p = n - p + 1$.Therefore,$p + q = n + 1$.

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

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