Question

The remainder when the sum ${\sum }_{n=1}^{100}(n^2+3n+1)$.$n!$ is divided by $10$ is

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is D $7$

${\sum }_{n=1}^{100}\left(\right(n+1\left)\right(n+2)-1)n!$
${\sum }_{n=1}^{100}\left(\right(n+2)!-n!)$
$=102!+101!-2!-1!$
$=102!+101!-3$
$=10k-3$
Remainder $=10-3=7$