Binomial Theorem And Its Simple Applications Question 166
Question: If $ x+y=1 $ , then $ \sum\limits_{r=0}^{n}{r^{2}{{,}^{n}}C_{r}x^{r}{y^{n-r}}} $ equals
Options:
A) nxy
B) $ nx(x+yn) $
C) $ nx(nx+y) $
D) None of these
Show Answer
Answer:
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) $
BETA
