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Sum of a series whose terms grow with constant second differences

This type of question, asked 5 times: 6 Apr 2023, 8 Apr 2023, 10 Apr 2023, 30 Jan 2024, 3 Apr 2025. All 5 questions below, oldest first.

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  1. Question 1

    6 Apr 2023

    68
    The sum of the first 20 terms of the series 5+11+5 + 11 + 19+29+41+…19 + 29 + 41 + \ldots is

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  2. Question 2

    8 Apr 2023

    152
    Let ana_{n} be the nth n^{\text{th~}} term of the series 5+8+14+235 + 8 + 14 + 23 +35+50+…+ 35 + 50 + \ldots and Sn=∑k=1nakS_{n}=\sum_{k = 1}^{n} a_{k}. Then S30−a40S_{30}-a_{40} is equal to

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  3. Question 3

    10 Apr 2023

    162
    If Sn=4+11+21+34+50+S_{n}= 4 + 11 + 21 + 34 + 50 + to nn terms, then 160( S29−S9)\frac{1}{60}\left( {\text{ }S}_{29}-S_{9} \right) is equal to

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  4. Question 4

    30 Jan 2024

    90
    Let α=12+42+82+132+192+262+……\alpha =1^{2}+4^{2}+8^{2}+13^{2}+19^{2}+26^{2}+ \ldots\ldots upto 10 terms and β=∑n=110n4\beta =\sum_{n = 1}^{10} n^{4}. If 4α−β=55k+404\alpha - \beta = 55k + 40, then kk is equal to

  5. Question 5

    3 Apr 2025

    57
    The sum 1+3+11+25+45+71+…1 + 3 + 11 + 25 + 45 + 71 + \ldots upto 20 terms, is equal to

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