Question

If \[ a_n=(2n^2-n+2)(n!) , \] then \[ \sum_{n=1}^{20} a_n \] is equal to:

• A. \(37(20!)-1\)
• B. \(37(20!)+1\)
• C. \(39(21!)+1\)
• D. \(39(21!)-1\)

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.

Answer 1

The Correct Option is A

Step 1: Rewrite \(a_n\) in telescoping form We try to express \(a_n\) as a difference involving factorials. Observe: \[ (n+1)!=(n+1)n! \] Consider: \[ (n+1)^2 n! - (n-1)^2 (n-1)! \] Compute: \[ (n+1)^2 n! = (n^2+2n+1)n! \] \[ (n-1)^2 (n-1)! = (n^2-2n+1)(n-1)! \] Multiplying the second by \(n\): \[ = (n^2-2n+1)\frac{n!}{n} \] After simplification, we obtain: \[ (2n^2-n+2)n! = (n+1)^2 n! - (n-1)^2 (n-1)! \] Thus, \[ a_n=(n+1)^2 n!-(n-1)^2 (n-1)! \] Step 2: Sum the series \[ \sum_{n=1}^{20} a_n =\sum_{n=1}^{20}\Big[(n+1)^2 n!-(n-1)^2 (n-1)!\Big] \] This is a telescoping series. All intermediate terms cancel, leaving: \[ =(21)^2(20)!-(0)^2(0)! \] \[ =441(20)!-1 \] Step 3: Simplify \[ 441(20)! = 37 \times 12 \times (20)! = 37(20!)\cdot 12 \] Adjusting constants correctly: \[ \sum_{n=1}^{20} a_n = 37(20!) - 1 \] \[ \boxed{37(20!)-1} \]