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Telescoping series: split each term and cancel to find the sum

This type of question, asked 3 times: CDS 2 2016, CDS 1 2021, CDS 1 2024. All 3 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    CDS 2 2016

    25
    What is 1+112+122+1+122+132+…+1+120072+120082\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \ldots + \sqrt{1 + \frac{1}{2007^2} + \frac{1}{2008^2}} equal to ?

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  2. Question 2

    CDS 1 2021

    30
    If 11×2+12×3+13×4+⋯+1n(n+1)=99100\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{n(n + 1)} = \frac{99}{100} then what is the value of nn ?

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  3. Question 3

    CDS 1 2024

    35
    What is 312×22+522×32+732×42+…\frac{3}{1^2 \times 2^2} + \frac{5}{2^2 \times 3^2} + \frac{7}{3^2 \times 4^2} + \ldots equal to ?

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