StatstriveISI intelligencePractice
Back to MSQE practice
PEAMCQEasy

2025 PEA Q17

Find ∑k=120241k(k+1)\sum_{k=1}^{2024}\dfrac{1}{k(k+1)}.

Reveal answer and solution

Answer

C

Solution

  1. 1

    Partial fractions:

  2. 2
    1k(k+1)=1k−1k+1. \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.
  3. 3

    Telescoping,

  4. 4
    ∑k=12024(1k−1k+1)=1−12025=20242025. \sum_{k=1}^{2024}\Big(\frac{1}{k} - \frac{1}{k+1}\Big) = 1 - \frac{1}{2025} = \frac{2024}{2025}.

Answer structure / marking notes

Count the index carefully: the last term is 12024⋅2025\frac{1}{2024\cdot 2025}, so the sum stops at 1−120251 - \frac{1}{2025}.

%==========================================

Content note

Imported from public/resources/isi/msqe/solutions/pea/2025/ISI_MSQE_PEA_2025_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.