A) \[np\]
B) \[nq\]
C) \[n(p+q)\]
D) None of these
Correct Answer: A
Solution :
\[\therefore \] \[\overline{x}\frac{...+n\cdot {{p}^{n}}]}{{{q}^{n}}+\frac{n}{1}{{q}^{n-1}}p+\frac{n(n-1)}{2}{{q}^{n-2}}{{p}^{2}}+...+{{p}^{n}}}\] \[=\frac{{{0.}^{n}}{{C}_{0}}{{q}^{n}}{{p}^{0}}+1\cdot {{\,}^{n}}{{C}_{1}}{{q}^{n-1}}p+...+{{n}^{n}}{{C}_{n}}{{q}^{0}}{{p}^{n}}}{^{n}{{C}_{0}}{{q}^{n}}{{p}^{0}}{{+}^{n}}{{C}_{1}}{{q}^{n-1}}{{p}^{1}}+...{{+}^{n}}{{C}_{n}}{{q}^{n-n}}{{p}^{n}}}\] \[=\frac{\sum\limits_{r\,=\,0}^{n}{r{{\cdot }^{n}}{{C}_{r}}{{q}^{n-r}}{{d}^{r}}}}{\sum\limits_{r\,=\,0}^{n}{^{n}{{C}_{r}}{{q}^{n-r}}{{p}^{r}}}}\] \[=\frac{\sum\limits_{r\,=\,0}^{n}{r\cdot \frac{n}{r}{{\,}^{n-1}}{{C}_{r-1}}{{q}^{n-r}}\,p\cdot {{p}^{r-1}}}}{\sum\limits_{r\,=\,0}^{n}{^{n}{{C}_{r}}{{q}^{n-r}}{{p}^{r}}}}\] \[=\frac{np{{(p+q)}^{n-1}}}{{{(q+p)}^{n}}}=np\] \[(\because \,\,p+q=1)\]
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