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I was solving some practice problems on recurrence relation for my upcoming exam and came across the following question.

Solve the recurrence relation T(n) = (n-1) T(n-1) + (n+1)! with the initial condition T(1) = 1.

I tried several techniques to solve it but it was of no use. Your help is very much appreciated.

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    $\begingroup$ This one is quite tricky. WolframAlpha outputs $\frac13(n+2)! - (n-1)!$ as the solution, but I'm not sure how to derive that... $\endgroup$ Commented Mar 17, 2021 at 21:46

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One thing you can do is to consider the Sequence $$U(n)=\frac{T(n)}{(n-1)!}$$ Hence you can show $$ U(n+1)=U(n)+(n+1)(n+2) $$ You can then, sum this relation to find $U(n)$ (telescopic sums)

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    $\begingroup$ This surely helps. Thanks for your suggestion. $\endgroup$
    – Kevin
    Commented Mar 17, 2021 at 22:01

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