Question

If n be a positive integer such that $n\ge 3$ then the value of the sum to n terms of $1\cdot n-\frac{(n-1)}{1!}(n-1)+\frac{(n-1)(n-2)}{2!}(n-2)-\frac{(n-1)(n-2)(n-3)}{3!}(n-3)+.......$ is:

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$0$

$S={\sum }_{r=0}^{n-1}{{}^{n-1}C}_r(n-r)({-1)}^r$

$=(-{1)}^r{\sum }_{r=0}^{n-1}{n{.}^{n-1}C}_r-{\sum }_{r=0}^{n-1}{r{.}^{n-1}C}_r({-1)}^r$

$\Rightarrow r.{{}^{n}C}_r={n{.}^{n-1}C}_{r-1}$

$\Rightarrow S=n{\sum }_{r=0}^{n-1}{(-{1)}^r{}^{n-1}C}_r$ $-\sum (n-1).({-1)}^r{{}^{n-2}C}_{r-1}$

Now, ${\sum }_{r=0}^{n-1}{(-{1)}^r{}^{n}C}_r=0$

$\therefore S=0-0$

$=0.$

Hence, the answer is $0.$