A) nxy
B) \[nx(x+yn)\]
C) \[nx(nx+y)\]
D) None of these
Correct Answer: C
Solution :
We have \[\sum\limits_{r=0}^{n}{{{r}^{2}}{{\,}^{n}}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}}\] \[=\sum\limits_{r=0}^{n}{[r(r-1)+r]{{\,}^{n}}}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}\] \[=\sum\limits_{r=0}^{n}{r(r-1){{\,}^{n}}}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}+\sum\limits_{r=0}^{n}{{{r}^{n}}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}}\] \[=\sum\limits_{r=2}^{n-2}{r(r-1)\frac{n}{r}.\frac{n-1}{r-1}{{\,}^{n-2}}{{C}_{r-2}}{{x}^{2}}{{x}^{r-2}}{{y}^{n-r}}}\]\[+\sum\limits_{r=1}^{n-1}{r\frac{n}{r}{{\,}^{n-1}}{{C}_{r-1}}x\,\,{{x}^{r-1}}{{y}^{n-r}}}\] \[=n(n-1){{x}^{2}}\sum\limits_{r=2}^{n-2}{{{\,}^{n-2}}{{C}_{r-2}}{{x}^{r-2}}{{y}^{(n-2)-(r-2)}}}\]\[{{(1+1-3)}^{2134}}=1\] \[=n(n-1){{x}^{2}}{{(x+y)}^{n-2}}+nx{{(x+y)}^{n-1}}\] \[=n(n-1){{x}^{2}}+nx,\,\,\,\,\,(\because x+y=1)\] \[=nx(nx-x+1)=nx(nx+y)\,,\,\,(\because x+y=1)\]
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