The sum sumlimits_{r = 1}^{10} {({r^2} + 1) t | vedclass

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(B) Let the general term be $T_r = (r^2 + 1)r!$. We can rewrite the term as: $T_r = (r^2 + r - r + 1)r! = (r(r+1) - (r-1))r!$ $T_r = r(r+1)! - (r-1)r!

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$10 	imes (11!)$

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(B) Let the general term be $T_r = (r^2 + 1)r!$. We can rewrite the term as: $T_r = (r^2 + r - r + 1)r! = (r(r+1) - (r-1))r!$ $T_r = r(r+1)! - (r-1)r!$ This is a telescoping series of the form $f(r) - f(r-1)$ where $f(r) = r(r+1)!$. Summing from $r=1$ to $10$: $\sum_{r=1}^{10} T_r = \sum_{r=1}^{10} [r(r+1)! - (r-1)r!] $ $= [1(2!) - 0(1!)] + [2(3!) - 1(2!)] + [3(4!) - 2(3!)] + \dots + [10(11!) - 9(10!)]$ $= 10(11!)$.

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

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