In the expansion of {(1 + x)^n} the coefficie | vedclass

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(b) term $ = {T_p} = {}^n{C_{p - 1}}{(x)^{n - p + 1}}{(1)^{p - 1}} = p$ 

$(p + 1)^{th}$ term $ = {T_{p + 1}} = {}^n{C_p}{(x)^{n - p}}{(1)^p} =

Vedclass · Accepted Answer

$n + 1$

Vedclass · Answer

(b) term $ = {T_p} = {}^n{C_{p - 1}}{(x)^{n - p + 1}}{(1)^{p - 1}} = p$ 

$(p + 1)^{th}$ term $ = {T_{p + 1}} = {}^n{C_p}{(x)^{n - p}}{(1)^p} = q$ 

Then, coefficient of $\frac{p}{q} = \frac{{{}^n{C_{p - 1}}}}{{{}^n{C_p}}}$ 

==> $\frac{p}{q} = \frac{{n!}}{{\left( {p - 1} \right)\,!\,\left( {n - p + 1}\right)\,\,!}}\,.\,\frac{{p\,!\,\,\,\left( {n - p} \right)\,\,!}}{{n\,!}}$

==> $\frac{p}{q} = \frac{p}{{n - p + 1}}$

==> $p + q = n + 1$.

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

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