The sum sum_{r=1}^{10} (r^2 + 1) times (r!) i | vedclass

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(B) Let the general term be $T_r = (r^2 + 1)r!$.We can rewrite the term as: $T_r = (r^2 + r - r + 1)r! = (r(r+1) - (r-1))r!$.This does not simplify ea

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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: $T_r = (r^2 + r - r + 1)r! = (r(r+1) - (r-1))r!$.This does not simplify easily,so let's use the form: $T_r = (r^2 + r - r + 1)r! = r(r+1)r! - (r-1)r!$.Note that $r(r+1)r! = r(r+1)!$.So,$T_r = r(r+1)! - (r-1)r!$.This is a telescoping series of the form $f(r) - f(r-1)$ where $f(r) = r(r+1)!$.Summing from $r=1$ to $10$:$\sum_{r=1}^{10} (r(r+1)! - (r-1)r!) = [1(2!) - 0(1!)] + [2(3!) - 1(2!)] + [3(4!) - 2(3!)] + \dots + [10(11!) - 9(10!)]$.All intermediate terms cancel out,leaving the last term: $10(11!)$.

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

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