If _r=1^10 r!(r^3+6r^2+2r+5) = (11!), then the value of is equal to: Answer: 160

Question

If $\displaystyle\sum_{r=1}^{10} r!(r^3+6r^2+2r+5) = \alpha(11!)$, then the value of $\alpha$ is equal to:

Answer: 160

Accepted Answer

Rewrite: $r!(r^3+6r^2+2r+5) = r!\left[(r+1)(r+2)(r+3) - 9(r+1) + 8\right]$
$$= (r+3)! - (r+1)! - 8\left[(r+1)! - r!\right]$$

Telescoping the sum from $r=1$ to $10$:
$$\sum = (13!+12!-2!-3!) - 8(11!-1!)$$
$$= (12\cdot13+12-8)\cdot11! - 8(11!-1) = 160\cdot11! - 8\cdot11! + 8\cdot11!... $$

After simplification, $\alpha = 160$.

Solutions · 1