A) \[nxy\]
B) \[nx(x+yn)\]
C) \[nx(n\,\,x+y)\]
D) None of these
Correct Answer: C
Solution :
\[\underset{r=0}{\overset{n}{\mathop{\Sigma }}}\,{{r}^{2}}\,{{\,}^{n}}{{C}_{r}}{{x}^{r}}\,{{y}^{n-r}}\] \[=\underset{r=0}{\overset{n}{\mathop{\Sigma }}}\,[r(r-1)+r]{{\,}^{n}}{{C}_{r}}\,{{x}^{r}}\,{{y}^{n-r}}\] \[=\underset{r=0}{\overset{n}{\mathop{\Sigma }}}\,r(r-1){{\,}^{n}}{{C}_{r}}\,{{x}^{r}}{{y}^{n-r}}\] \[+\underset{r=0}{\overset{n}{\mathop{\Sigma }}}\,{{r}^{n}}{{C}_{r}}\,{{x}^{r}}\,{{y}^{n-r}}\] \[=\underset{r=2}{\overset{n-r}{\mathop{\Sigma }}}\,r(r-1)\frac{n}{r}\cdot \frac{n-1}{r-1}\] \[^{n-2}{{C}_{r-2}}{{x}^{2}}\cdot {{x}^{r-2}}{{y}^{n-r}}\] \[+\underset{r=1}{\overset{n}{\mathop{\Sigma }}}\,r\cdot {{\frac{n}{r}}^{n-1}}{{C}_{r-1}}x\cdot {{x}^{r-1}}{{y}^{n-r}}\] \[=n(n-1){{x}^{2}}\underset{r=2}{\overset{n-2}{\mathop{\Sigma }}}\,{{\,}^{n-2}}{{C}_{r-2}}\] \[{{x}^{r-2}}{{y}^{(n-2)-(r-2)}}\] \[+nx\sum\limits_{r=1}^{n}{^{n-1}{{C}_{r-1}}}{{x}^{r-1}}{{y}^{(n-1)-(r-1)}}\] \[=n(n-1){{x}^{2}}{{(x+y)}^{n-2}}+nx{{(x+y)}^{n-1}}\] \[=n(n-1){{x}^{2}}+nx\] \[(\because \,\,x+y=1)\] \[=nx(nx-x+1)\] \[=nx(nx+y)\] \[(\because \,\,x+y=1)\]You need to login to perform this action.
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