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 $=\left(n+2\right)2^{n+1}\sum {\left(-1\right)}^r\left(\_{}^n{C}_r^{}\right){\left(\frac{1}{2}\right)}^r+2^{n}n\sum {\left(-1\right)}^{r-1}\left(\_{}^{n-1}{C}_{r-1}^{}\right){\left(\frac{1}{2}\right)}^{r-1}$
$=\left(n+2\right)2^{n+1}{\left(1-\frac{1}{2}\right)}^n+2^{n}n{\left(1-\frac{1}{2}\right)}^{n-1}$
$=\left(n+2\right)2^{n+1}\frac{1}{2^n}+2^{n}n\frac{1}{2^{n-1}}$
$=\left(n+2\right)2+2n$
$=4n+4=4\left(n+1\right)$