Question

The sum $\frac{1}{1 !(n-1) !}+\frac{1}{2 !(n-2) !}+\ldots \ldots \frac{1}{1 !(n-1) !}$ is equal to :

• A. $\frac{1}{n !}\left(2^{n-1}-1\right)$
• B. $\frac{2}{n !}\left(2^n-1\right)$
• C. $\frac{2}{n !}\left(2^{n-1}-1\right)$
• D. none

Answer 1

Answer: (C)

$\frac{1}{1 !(n-1) !}+\frac{1}{2 !(n-2) !}+\frac{1}{1 !(n-1) !} \ldots+=\frac{1}{n !}\left[{ }^n C_1+{ }^n C_2+\ldots .+{ }^n C_{n-1}\right]$ (multiply and divde by $n !$ )
$=\frac{1}{n !}\left[2^n-2\right]=\frac{2}{n !}\left(2^{n-1}-1\right)$