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Telescoping Sum 528/[n(n+1)(n+2)] = 130 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
∑n=110(528n(n+1)(n+2))\displaystyle\sum_{n=1}^{10}\left(\dfrac{528}{n(n+1)(n+2)}\right) is equal to
A6565
B130130correct
C220220
D440440
Solution
Step 1: 1n(n+1)(n+2)=12(1n(n+1)−1(n+1)(n+2))\dfrac{1}{n(n+1)(n+2)}=\dfrac12\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right). Step 2: Let an=1n(n+1)a_n=\dfrac{1}{n(n+1)}, so the sum telescopes:
∑n=1101n(n+1)(n+2)=12(a1−a11)=12(11⋅2−111⋅12)=12(12−1132).\sum_{n=1}^{10}\frac{1}{n(n+1)(n+2)}=\frac12\big(a_1-a_{11}\big)=\frac12\left(\frac{1}{1\cdot2}-\frac{1}{11\cdot12}\right)=\frac12\left(\frac12-\frac{1}{132}\right).
=12⋅66−1132=12⋅65132=65264.=\frac12\cdot\frac{66-1}{132}=\frac12\cdot\frac{65}{132}=\frac{65}{264}.
 Step 3: 528=2⋅264528=2\cdot264:
528⋅65264=2⋅65=130.528\cdot\frac{65}{264}=2\cdot65=130.
 Correct answer: (2)
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