Question:medium
If
\[
a_n=(2n^2-n+2)(n!) ,
\]
then
\[
\sum_{n=1}^{20} a_n
\]
is equal to:
If
\[
a_n=(2n^2-n+2)(n!) ,
\]
then
\[
\sum_{n=1}^{20} a_n
\]
is equal to:
Show Hint
Whenever factorial terms appear with polynomials in \(n\),
try expressing the term as a difference of successive factorial expressions —
this often leads to telescoping sums.
Updated On: Mar 3, 2026
- \(37(20!)-1\)
- \(37(20!)+1\)
- \(39(21!)+1\)
- \(39(21!)-1\)
Show Solution
The Correct Option is A
Solution and Explanation
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