Question

The value of $\left(n + 2\right)\cdot \_{}^{n}C_{0}^{}\cdot 2^{n + 1}-\left(n + 1\right)\cdot \_{}^{n}C_{1}^{}\cdot 2^{n}+\left(n\right)\cdot \_{}^{n}C_{2}^{}\cdot 2^{n - 1}-....$ to $\left(n + 1\right)$ terms is equal to

• A. $4$
• B. $4n$
• C. $4\left(n + 1\right)$
• D. $2\left(n + 2\right)$

Answer 1

Answer: (C)

$T_{r + 1}=\left(- 1\right)^{r}\left(n - r + 2\right)\_{}^{n}C_{r}^{}2^{n - r + 1}$
$=(n+2) 2^{n+1}(-1)^{r}\left({ }^{n} C_{r}\right)\left(\frac{1}{2}\right)^{r}-2^{n+1}(-1)^{r} r\left({ }^{n} C_{r}\right)\left(\frac{1}{2}\right)^{r}$
$=(n+2) 2^{n+1}(-1)^{r}\left({ }^{n} C_{r}\right)\left(\frac{1}{2}\right)^{r}+2^{n} \cdot n(-1)^{r-1}\left({ }^{n-1} C_{r-1}\right)\left(\frac{1}{2}\right)^{r-1}$
So, sum $=\left(n + 2\right)2^{n + 1}\displaystyle \sum \left(- 1\right)^{r}\left(\_{}^{n}C_{r}^{}\right)\left(\frac{1}{2}\right)^{r}+2^{n}n\displaystyle \sum \left(- 1\right)^{r - 1}\left(\_{}^{n - 1}C_{r - 1}^{}\right)\left(\frac{1}{2}\right)^{r - 1}$
$=\left(n + 2\right)2^{n + 1}\left(1 - \frac{1}{2}\right)^{n}+2^{n}n\left(1 - \frac{1}{2}\right)^{n - 1}$
$=\left(n + 2\right)2^{n + 1}\frac{1}{2^{n}}+2^{n}n\frac{1}{2^{n - 1}}$
$=\left(n + 2\right)2+2n$
$=4n+4=4\left(n + 1\right)$