Question

Let $t_n=n.(n!)$. Then ${\sum }_{n=1}^{15}{t}_n$ is equal to

• A. $15!-1$
• B. $15!+1$
• C. $16!-1$
• D. none of these

Answer 1

The correct option is C $16!-1$

Given that $t_n=n.(n!)=(n+1)n!-n!$

$\Rightarrow t_n=(n+1)!-n!$
Now, $S={\sum }_{n=1}^{15}{t}_n={\sum }_{n=1}^{15}[(n+1)!-n!]$

$\Rightarrow S=(2!-1!)+(3!-2!)+(4!-3!)+.....+(16!-15!)$

$\Rightarrow S=16!-1$