Question

If $n$is an integer greater than $1,$then $a-{}^{n}C_1(a-1)+{}^{n}C_2(a-2)-.....+{(-1)}^n(a-n)$

• A. $a$
• B. $0$
• C. $a^2$
• D. $2^n$

Answer 1

The correct option is B

$0$

Explanation for the correct option:

$a-{}^{n}C_1(a-1)+{}^{n}C_2(a-2)-.....+{(-1)}^n(a-n)$

$$\begin{gathered} =\sum _{r=0}^n(-1){}^{rn}C_r(a-r) \\ =\sum _{r=0}^n(-1){}^{rn}C_r·a-\sum _{r=0}^n(-1){}^{rn}C_r·r \end{gathered}$$

$=a\sum _{r=0}^n(-1){}^{rn}C_r-\sum _{r=1}^n(-1){}^{rn}C_r·r$ (when $r=0,$ we get $0$)

$$\begin{gathered} =a{(1-1)}^n-\sum _{r=1}^n(-1){}^r\frac{n}{r}{}^{n-1}C_{r-1}·r\left[\because {}^{n}C_r=\frac{n}{r}{}^{n-1}C_{r-1}\right] \\ =a\left(0\right)-\sum _{r=1}^n(-1){}^{r}n{}^{n-1}C_{r-1} \\ =0-n\sum _{r=1}^n(-1){}^r{}^{n-1}C_{r-1} \\ =-n\left[-{}^{n-1}C_0+{}^{n-1}C_1-{}^{n-1}C_2.....{(-1)}^{n-1}{}^{n-1}C_{r-1}\right] \\ =-n\left[0\right]\left[\because {(1-1)}^{n-1}={}^{n-1}C_0-{}^{n-1}C_1+{}^{n-1}C_2-{}^{n-1}C_3+....={(-1)}^n{}^{n-1}C_{n-1}=0\right] \\ =0 \end{gathered}$$

Hence the correct option is option(b) i.e. $0.$