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Sums involving r! that telescope

This type of question, asked 2 times: 2021 Paper 10, 29 Jul 2022. All 2 questions below, oldest first.

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  1. Question 1

    2021 Paper 10

    83
    If ∑r=110r!(r3+6r2+2r+5)=α(11!)\sum_{r = 1}^{10} r!\left( r^{3}+ 6r^{2}+ 2r + 5 \right)=\alpha(11!), then the value of α\alpha is equal to

  2. Question 2

    29 Jul 2022

    157
    ∑r=120(r2+1)(r!)\sum_{r = 1}^{20} \left( r^{2}+ 1 \right)(r!) is equal to:

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