Question

Let the sum of the series
$(1^2+1)!+(2^2+1)2!+(3^2+1)3!+...+(n^2+1)n!$ be $n(n+k)!$. Find $k+4$ ?

Answer 1

Let $S_n=(1^2+1)!+(2^2+1)2!+(3^2+1)3!+...+(n^2+1)n!$
$\therefore nthterm{T}_n=(n^2+1)n!$
$=\left\{\right(n+1\left)\right(n+2)-3(n+1)+2\}n!$
$T_n=(n+2)!-3(n+1)!+2n!$
Putting $n=1,2,3,4,...,n$
Then,
$T_1=3!-3.2!+2.1!$
$T_2=4!-3.3!+2.2!$
$T_3=5!-3.4!+2.3!$
$T_4=6!-3.5!+2.4!$
$........................$
$........................$
$T_{n-1}=(n+1)!-3n!+2(n-1)!$
$T_n=(n+2)!-3(n+1)!+2\left(n\right)!$
$\therefore S_n=T_1+T_2+T_3+...+T_n$
$=(n+2)!-2(n+1)!\left(\text{the rest cancels out}\right)$
$=(n+2)(n+1)!-2(n+1)!$
$=(n+1)!(n+2-2)$
$=n(n+1)!$
$\therefore k=1$
$k+4=5$