The sum $$\sum_{r=1}^{10} (r^2 + 1) \times (r!)$$, is equal to:

We have to evaluate the finite sum

$$S=\sum_{r=1}^{10}\bigl(r^{2}+1\bigr)\,r!.$$

To simplify every term we try to express $$(r^{2}+1)r!$$ as the difference of two neighbouring factorial expressions, so that a telescoping effect will appear in the sum.

Let us start from the identity for factorials

$$ (r+1)!=(r+1)\,r!. $$

We multiply both sides by an integer $$k$$ (to be chosen in a moment) and write

$$ k(r+1)!-k r! = k\bigl[(r+1)\,r!-r!\bigr]=k\bigl(r\,r!\bigr). $$

If we now choose the integer $$k$$ to be $$r$$, the right-hand side becomes

$$ r\bigl(r\,r!\bigr)=r^{2}r!. $$

Thus we obtain

$$ r(r+1)!-r\,r! = r^{2}r!. $$

We still need the extra “$$+1$$” which sits beside $$r^{2}$$ in $$(r^{2}+1)r!$$. We introduce it by subtracting $$(r-1)r!$$ on the right and on the left:

$$ r(r+1)!-\bigl(r-1\bigr)r! = r^{2}r!+r!-(r-1)r! = (r^{2}+1)r!. $$

Hence we have established the very convenient relation

$$ (r^{2}+1)r! = r(r+1)!-\bigl(r-1\bigr)r!. \cdots(1) $$

Now we go back to the required sum $$S$$ and substitute the expression (1) for every term:

$$ \begin{aligned} S &= \sum_{r=1}^{10}\Bigl[r(r+1)!-\bigl(r-1\bigr)r!\Bigr] \\ &= \sum_{r=1}^{10} r(r+1)! - \sum_{r=1}^{10}\bigl(r-1\bigr)r!. \cdots(2) \end{aligned} $$

Observe that the two sums in (2) are almost identical, only the indices differ slightly. To see the cancellation clearly, let us write out the first few and the last few terms of each:

First sum: $$\;1\cdot2! + 2\cdot3! + 3\cdot4! + \dots + 9\cdot10! + 10\cdot11!\,.$$
Second sum: $$0\cdot1! + 1\cdot2! + 2\cdot3! + 3\cdot4! + \dots + 8\cdot9! + 9\cdot10!\,.$$

Term by term everything cancels except the very first non-matching piece $$0\cdot1!$$ (which is zero anyway) and, crucially, the last piece $$10\cdot11!$$ from the first sum, which has no counterpart in the second sum.

So on subtracting, every term except $$10\cdot11!$$ disappears, giving

$$ S = 10\cdot11!\,. $$

We may leave it like this, but evaluating once more confirms the numerical value:

$$ 11! = 39\,916\,800,\quad\text{so}\quad S = 10\times39\,916\,800 = 399\,168\,000. $$

Among the given options, Option B states exactly $$10\times11!,$$ which is what we have obtained.

Hence, the correct answer is Option B.