PYQ Vault
EbooksTutor

Sum of series with factorials in the denominator, in terms of e

This type of question, asked 6 times: 2021 Paper 24, 2021 Paper 26, 30 Jan 2023, 1 Feb 2023, 3 Apr 2025, 28 Jan 2026 S1. All 6 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 24

    75
    Let Sn=1⋅(n−1)+2⋅(n−2)+3⋅(n−3)+….+S_{n}= 1 \cdot (n- 1) + 2 \cdot (n- 2) + 3 \cdot (n- 3) + \ldots. + (n−1)⋅1,n≥4(n - 1) \cdot 1,n \geq 4. The sum ∑n=4∞(2Snn!−1(n−2)!)\sum_{n= 4}^{\infty} \left( \frac{2S_{n}}{n!}-\frac{1}{(n- 2)!} \right) is equal to:

    Tap an option to check your answer.

  2. Question 2

    2021 Paper 26

    62
    The sum of the series ∑n=1∞n2+6n+10(2n+1)!\sum_{n= 1}^{\infty} \frac{n^{2}+ 6n+ 10}{(2n+ 1)!} is equal to:

    Tap an option to check your answer.

  3. Question 3

    30 Jan 2023

    84
    Let ∑n=0∞n3((2n)!)+(2n−1)(n!)(n!)((2n)!)=ae+be+c\sum_{n = 0}^{\infty} \frac{n^{3}((2n)!) + (2n - 1)(n!)}{(n!)((2n)!)}= ae +\frac{b}{e}+ c, where a,b,c∈Za,b,c \in Z and e=∑n=0∞1n!e =\sum_{n = 0}^{\infty} \frac{1}{n!} Then a2−b+ca^{2}- b + c is equal to

  4. Question 4

    1 Feb 2023

    152
    The sum ∑n=1∞2n2+3n+4(2n)!\sum_{n = 1}^{\infty} \frac{2n^{2}+ 3n + 4}{(2n)!} is equal to:

    Tap an option to check your answer.

  5. Question 5

    3 Apr 2025

    145
    The sum 1+1+32!+1+3+53!+1+3+5+74!+…1 +\frac{1 + 3}{2!}+\frac{1 + 3 + 5}{3!}+\frac{1 + 3 + 5 + 7}{4!}+ \ldots upto ∞\infty terms, is equal to

    Tap an option to check your answer.

  6. Question 6

    28 Jan 2026 S1

    52
    The value of ∑k=1∞(−1)k+1(k(k+1)k!)\sum_{k= 1}^{\infty} ( - 1)^{k+ 1}\left( \frac{k(k+ 1)}{k!} \right) is :

    Tap an option to check your answer.