Question

If $x+y=1$, then $\displaystyle\sum_{r=0}^{n} r^{2 \cdot n} C_{r} x^{r} y^{n-r}$ is equal to

• A. nxy
• B. nx (x +yn)
• C. nx (n x +y)
• D. None of these

Answer 1

Answer: (C)

$\displaystyle \sum_{r=0}^{n} r^{2}{ }^{n} C_{r} x^{r} y^{n-r}$
$=\displaystyle \sum_{r=0}^{n}[r(r-1)+r]{ }^{n} C_{r} x^{r} y^{n-r}$
$=\displaystyle \sum_{r=0}^{n} r(r-1){ }^{n} C_{r} x^{r} y^{n-r}$
$+\displaystyle \sum_{r=0}^{n} r^{n} C_{r} x^{r} y^{n-r}$
$=\displaystyle \sum_{r=2}^{n-2} r(r-1) \frac{n}{r} \cdot \frac{n-1}{r-1}{ }^{n-2} C_{r-2} x^{2} \cdot x^{r-2} y^{n-r}$
$+\displaystyle \sum_{r=1}^{n} r \cdot \frac{n}{r}{ }^{n-1} C_{r-1} x \cdot x^{r-1} y^{n-r}$
$=n(n-1) x^{2} \displaystyle \sum_{r=2}^{n-2} {}^{n-2} {C_{r-2}} x^{r-2} y^{(n-2)-(r-2)}$
$+n x \displaystyle\sum_{r=1}^{n}{ }^{n-1} C_{r-1} x^{r-1} y^{(n-1)-(r-1)}$
$=n(n-1) x^{2}(x +y)^{n-2}+n x(x +y)^{n-1}$
$=n(n-1) x^{2}+n x (\because x +y=1)$
$=n x(n x-x+1)$
$=n x(n x +y) (\because x +y=1)$