A) 0
B) 1
C) \[\frac{1}{4}\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
Given, \[{{(p+q)}^{n}}\] \[{{T}_{r}}={{T}_{(r+1)+1}}={{\,}^{n}}{{C}_{r-1}}{{p}^{n-r+1}}.{{q}^{r-1}}\] and \[{{T}_{r+1}}={{\,}^{n}}{{C}_{r}}{{p}^{n-r}}.{{q}^{r}}\] From question, \[^{n}{{C}_{r-1}}{{p}^{n-r+1}}.{{q}^{r-1}}={{\,}^{n}}{{C}_{r}}{{p}^{n-r}}.{{q}^{r}}\] \[\frac{n!}{(r-1)!(n-r+1)(n-r)!}.{{p}^{n-r}}{{q}^{r}}.\frac{p}{q}\] \[=\frac{n!}{r(r-1)(n-r)!}.{{p}^{n-r}}.{{q}^{r}}\] \[\Rightarrow \]\[\frac{1}{(n-r+1)}.\frac{p}{q}=\frac{1}{r}\] \[\Rightarrow \]\[pr=qn-qr+q\] \[\Rightarrow \]\[\frac{q(n+1)}{r(p+q)}=1\]
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