Question

The sum $\sum _{k=1}^{10}k.k!$ equals

• A. $\left(10\right)!$
• B. $\left(11\right)!$
• C. $\left(10\right)!+1$
• D. $\left(11\right)!-1$

Answer 1

The correct option is D $\left(11\right)!-1$

$\sum _{k=1}^{10}k.k!$

$\sum _{k=1}^{10}\left[\left(k+1\right)k!-k!\right]$

$\sum _{k=1}^{10}\left(\left(k+1\right)!-k!\right)$

$\left(2!-1!\right)+\left(3!-2!\right)+\left(4!-3!\right)+\left(10!-9!\right)+\left(11!-10!\right)$

$=11!-1$