Question

The Sum $\sum_{r=1}^{20}(r^{2}+1)\times r!$ is equal to

• A. $21\times21!$
• B. $20\times21!$
• C. 21!
• D. $20\times21\times20!$

Hint:

The "Method of Differences" is the most powerful tool for factorial sums. Always look for a way to write the coefficient as $(r+k) - (\dots)$ to shift the index of the factorial.

Answer 1

The Correct Option is B

To evaluate this summation, we need to transform the general term into a format that allows the series to telescope.
Step 1: The general term is $T_r = (r^2 + 1)r!$. We seek to express the polynomial $(r^2 + 1)$ in terms of $r$ and $(r+1)$ to link it with the factorial. Observe that: $r^2 + 1 = (r^2 + r) - (r - 1) = r(r+1) - (r-1)$. This allows us to write: $T_r = [r(r+1) - (r-1)]r!$.
Step 2: Distribute $r!$ across the brackets: $T_r = r(r+1)r! - (r-1)r!$ Recall the property of factorials: $(r+1)r! = (r+1)!$. Substituting this into the first part of our expression: $T_r = r(r+1)! - (r-1)r!$.
Step 3: Let us define a sequence $V_r = r(r+1)!$. Then, it follows that $V_{r-1} = (r-1)(r-1+1)! = (r-1)r!$. Our general term is now beautifully simplified to $T_r = V_r - V_{r-1}$.
Step 4: The sum $S = \sum_{r=1}^{20} T_r$ is: $S = (V_1 - V_0) + (V_2 - V_1) + (V_3 - V_2) + \dots + (V_{20} - V_{19})$. This is a telescoping sum where every intermediate term cancels out. The result is $S = V_{20} - V_0$. $V_{20} = 20(20+1)! = 20 \times 21!$. $V_0 = 0(0+1)! = 0$. Thus, $S = 20 \times 21!$. The final result matches Option (2).