Question

Find sum of the series$\sum _{r=1}^n(r^2+1)(r!)=$

• A. $(n+1)!$
• B. $(n+2)!-1$
• C. $n(n+1)!$
• D. $noneofthese!$

Answer 1

The correct option is C $n(n+1)!$

Consider the $i^{th}$ term: $t_i=(i^2+1)i!$.
We re-write it as :
$t_i=\left(\right(i+1\left)\right(i+2)-3(i+1)+2)i!=(i+2)!-3(i+1)!+2i!$
$t_1=3!-3.2!+2.1!$
$t_2=4!-3.3!+2.2!$
$t_3=5!-3.4!+2.3!$
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.
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$t_{n-2}=n!-3.(n-1)!+2.(n-2)!$
$t_{n-1}=(n+1)!-3.n!+2.(n-1)!$
$t_n=(n+2)!-3.(n+1)!+2.n!$
Observe that diagonal terms(from top left to bottom right) having similar factorials and get cancelled when added together. Remaining terms are collected.
Answer = $2.1!-2!+(n+2)!-2.(n+1)!=n(n+1)!$