JEE Mains · Maths · STD 11 - 8. sequence and series
\(\displaystyle\sum_{n=1}^{10} \left( \dfrac{528}{n(n+1)(n+2)} \right)\) is equal to:
- A \(65\)
- B \(130\)
- C \(220\)
- D \(440\)
Answer & Solution
Correct Answer
(B) \(130\)
Step-by-step Solution
Detailed explanation
Let \(T_n = \dfrac{528}{n(n+1)(n+2)}\) \(T_n = \dfrac{528}{2} \left( \dfrac{(n+2) - n}{n(n+1)(n+2)} \right)\) \(T_n = 264 \left( \dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)} \right)\) The sum of the first \(10\) terms is given by:…
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