Evaluating the Summation $\sum_{r=1}^{20} (r^2+1)\times r!$

We need to calculate the sum $S = \sum_{r=1}^{20} (r^2+1) r!$. To solve this, we can rewrite the term $(r^2+1) r!$ in a way that allows for a telescoping sum.

Rewriting the General Term

Consider the general term $(r^2+1) r!$. We can express $r^2+1$ as $(r^2+r+1) - r$. So, the term becomes:

$(r^2+1) r! = ((r^2+r+1) - r) r! = (r^2+r+1) r! - r \cdot r!$

We know the identity $r \cdot r! = (r+1)! - r!$. Also, notice that $(r^2+r+1) r! = (r+1)r \cdot r! + r! = (r+1)(r+1)! - r \cdot r!$. Therefore, we can rewrite the general term as:

$(r^2+1) r! = [(r+1)(r+1)! - r \cdot r!] - r \cdot r!$

This simplifies to:

$(r^2+1) r! = (r+1)(r+1)! - 2 (r \cdot r!)$

Substitute $r \cdot r! = (r+1)! - r!$:

$(r^2+1) r! = (r+1)(r+1)! - 2 ((r+1)! - r!)$

$(r^2+1) r! = (r+1)(r+1)! - 2(r+1)! + 2r!$

$(r^2+1) r! = (r+1-2)(r+1)! + 2r! = (r-1)(r+1)! + 2r!$. This manipulation seems complex.

Let's try another approach using the identity $f(r+1) - f(r)$. Let $f(r) = r \cdot r!$. Then $f(r+1) = (r+1)(r+1)!$. $f(r+1) - f(r) = (r+1)(r+1)! - r \cdot r! = (r+1)(r+1)r! - r \cdot r!$ $= [(r+1)^2 - r] r! = (r^2+2r+1-r) r! = (r^2+r+1) r!$.

The general term is $(r^2+1)r!$. We can write this as:

$(r^2+1) r! = (r^2+r+1) r! - r \cdot r!$

Using $f(r) = r \cdot r!$, we have $(r^2+1) r! = [f(r+1) - f(r)] - r \cdot r!$.

Applying Telescoping Summation

The sum $S$ can be written as:

$S = \sum_{r=1}^{20} [(f(r+1) - f(r)) - r \cdot r!]$

$S = \sum_{r=1}^{20} (f(r+1) - f(r)) - \sum_{r=1}^{20} r \cdot r!$

The first part is a telescoping sum:

$\sum_{r=1}^{20} (f(r+1) - f(r)) = f(21) - f(1)$

$= 21 \cdot 21! - 1 \cdot 1! = 21 \cdot 21! - 1$

The second part is also a telescoping sum using the identity $r \cdot r! = (r+1)! - r!$:

$\sum_{r=1}^{20} r \cdot r! = \sum_{r=1}^{20} ((r+1)! - r!)$

$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + (21! - 20!)$

$= 21! - 1! = 21! - 1$

Final Calculation

Now, substitute the results of the two sums back into the expression for S:

$S = (21 \cdot 21! - 1) - (21! - 1)$

$S = 21 \cdot 21! - 1 - 21! + 1$

$S = 21 \cdot 21! - 21!$

$S = (21 - 1) \cdot 21!$

$S = 20 \cdot 21!$

Thus, the sum $\sum_{r=1}^{20} (r^2+1)\times r!$ is equal to $20 \times 21!$.