Question

Value of
$S={{}^{n}C}_r+3\left({{}^{n-1}C}_r\right)+5\left({{}^{n-2}C}_r\right)+...+$ upto $(n-r+1)terms$

• A. ${{}^{n+2}C}_{r+2}$
• B. ${{}^{n+2}C}_{r+2}+{{}^{n+1}C}_{r+2}$
• C. ${{}^{n+2}C}_{r+1}$
• D. ${{}^{n+2}C}_{r+2}+{{}^{n+1}C}_r$

Answer 1

The correct option is B ${{}^{n+2}C}_{r+2}+{{}^{n+1}C}_{r+2}$

$S=$ coefficient of $x^r$ in $E$ where $E={(1+x)}^n+3{(1+x)}^{n-1}+5{(1+x)}^{n-2}+...+\left(2\right(n-r)+1)({(1+x)}^r$
Let $F=a^n+3a^{n-1}+.....\left(2\right(n-r)+1)a^r$
$\Rightarrow \frac{1}{a}F=a^{n-1}+....+\left(2\right(n-r)-3)a^r+\left(2\right(n-r)+1)a^{r-1}$
$\Rightarrow (1-\frac{1}{a})F=a^n+2a^{n-1}+....+2a^r-\left(2\right(n-r)+1)a^{r-1}$
$=a^n+\frac{2a^r(1-a^{n-r})}{1-a}-\left(2\right(n-r)+1)a^{r-1}$
$\Rightarrow F=\frac{a^{n+1}}{a-1}-\frac{2a^{r+1}(1-a^{n-r})}{{(1-a)}^2}-\frac{\left(2\right(n-r)+1)a^r}{a-1}$
Thus,
$E=\frac{{(1+x)}^{n+1}}{x}+\frac{2{(1+x)}^{n+1}}{x^2}-\frac{2{(1+x)}^{r+1}}{x^2}-\frac{\left(2\right(n-r)+1){(1+x)}^r}{x}$
$\therefore S={{}^{n+1}C}_{r+1}+2\left({{}^{n+1}C}_{r+2}\right)={{}^{n+2}C}_{r+2}+\left({{}^{n+1}C}_{r+2}\right)$
Hence, option B.