A) np2
B) npq
C) np
D) None of these
Correct Answer: C
Solution :
We have, \[\sum\limits_{r=0}^{n}{{{r}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}\] \[=\sum\limits_{r=0}^{n}{r}.{{\frac{n}{r}}^{n-1}}{{C}_{r-1}}p.{{p}^{r-1}}{{q}^{(n-1)-(r-1)}}\] \[=np\left\{ \sum\limits_{r=0}^{n}{^{n-1}}{{C}_{r-1}}{{p}^{r-1}}{{q}^{(n-1)-(r-1)}} \right\}\] \[=np{{(q+p)}^{n-1}}\left[ \because {{(q+q)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}} \right]\] \[=np\] \[(\because p+q+1)\]You need to login to perform this action.
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