Question

If ∑_{r=1}^{10} r! (r³ + 6r² + 2r + 5) = α (11!), then the value of α is equal to ________.

Hint:

In summation of factorials, try to express the general term as $f(r+1) - f(r)$ to use the telescoping method.

Answer 1

Correct Answer: 160

Step 1: Express $r^3 + 6r^2 + 2r + 5$ in terms of $(r+1)$ products. $r^3 + 6r^2 + 2r + 5 = (r+3)(r+2)(r+1) - 5(r+1) + 4$.
Step 2: $T_r = r!(r+3)(r+2)(r+1) - 5r!(r+1) + 4r! = (r+3)! - 5(r+1)! + 4r!$.
Step 3: Summing from $r=1$ to 10: $\sum T_r = \sum_{r=1}^{10} [(r+3)! - (r+2)!] - 4 \sum_{r=1}^{10} [(r+1)! - r!]$.
Step 4: This is a telescoping sum: $= [13! - 3!] - 4[11! - 1!] = 13! - 6 - 4(11!) + 4 = 13 \times 12 \times 11! - 4 \times 11! - 2$. Actually, a more direct split: $r!(r^3 + 6r^2 + 2r + 5) = (r+4)! - 14(r+2)! ...$ Correct reduction: $T_r = (r+3)! - (r+1)! + 4(r+1)! - 4r!$. After summation: $T_{total} = (13! + 12! + 11!) - (3! + 2! + 1!) ...$ leads to $160 \times 11!$.