JEE Main 2026 Apr 5 Shift 1, Mathematics Q5: Summation of Series

JEE Main2026Apr 5, Shift 1Mathematics

$n = 1 \sum 10 (\frac{528}{n ( n + 1 ) ( n + 2 )})$ is equal to :

A
65
B
130
C
220
D
440

Solution

Use the telescoping identity $\frac{1}{n ( n + 1 ) ( n + 2 )} = \frac{1}{2} (\frac{1}{n ( n + 1 )} - \frac{1}{( n + 1 ) ( n + 2 )})$ .

Summing from $n = 1$ to $10$ collapses to $\frac{1}{2} (\frac{1}{1 \cdot 2} - \frac{1}{11 \cdot 12}) = \frac{1}{2} \cdot \frac{65}{132} = \frac{65}{264}$ .

Multiplying by $528$ : $\frac{65}{264} \times 528 = 130$ .

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JEE Main 2026 Apr 5 Shift 1, Mathematics Q5: Summation of Series

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