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JEE Main 20265 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
5 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 5 April · Morning Shift · Mathematics · Sequences and Series

∑_ n=1 ^ 10 ( (528)/(n(n+1)(n+2)) ) is equal to:

\displaystyle\sum_{n=1}^{10} (528n(n+1)(n+2))\left( \dfrac{528}{n(n+1)(n+2)} \right) is equal to:

Options

  1. A6565
  2. B130130
  3. C220220
  4. D440440

Correct answer

B. 130130

Step-by-step solution

Let Tn=528n(n+1)(n+2)T_n = \dfrac{528}{n(n+1)(n+2)}

Tn=5282((n+2)−nn(n+1)(n+2))T_n = \dfrac{528}{2} \left( \dfrac{(n+2) - n}{n(n+1)(n+2)} \right)

Tn=264(1n(n+1)−1(n+1)(n+2))T_n = 264 \left( \dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)} \right)

The sum of the first 1010 terms is given by:

S10=∑n=110Tn=264∑n=110(1n(n+1)−1(n+1)(n+2))S_{10} = \displaystyle\sum_{n=1}^{10} T_n = 264 \displaystyle\sum_{n=1}^{10} \left( \dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)} \right)

This is a telescoping series, so most terms cancel out:

S10=264((11⋅2−12⋅3)+(12⋅3−13⋅4)+⋯+(110⋅11−111⋅12))S_{10} = 264 \left( \left( \dfrac{1}{1 \cdot 2} - \dfrac{1}{2 \cdot 3} \right) + \left( \dfrac{1}{2 \cdot 3} - \dfrac{1}{3 \cdot 4} \right) + \dots + \left( \dfrac{1}{10 \cdot 11} - \dfrac{1}{11 \cdot 12} \right) \right)

S10=264(12−111⋅12)S_{10} = 264 \left( \dfrac{1}{2} - \dfrac{1}{11 \cdot 12} \right)

S10=264(12−1132)S_{10} = 264 \left( \dfrac{1}{2} - \dfrac{1}{132} \right)

S10=264(66−1132)S_{10} = 264 \left( \dfrac{66 - 1}{132} \right)

S10=264×65132S_{10} = 264 \times \dfrac{65}{132}

S10=2×65=130S_{10} = 2 \times 65 = 130

Answer: 130130
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