JEE Mains · Maths · STD 11 - 8. sequence and series
The sum \(1 + \dfrac{1}{2}(1^2 + 2^2) + \dfrac{1}{3}(1^2 + 2^2 + 3^2) + \ldots\) upto 10 terms is equal to :
- A \(130\)
- B \(155\)
- C \(\dfrac{315}{2}\)
- D \(\dfrac{325}{2}\)
Answer & Solution
Correct Answer
(C) \(\dfrac{315}{2}\)
Step-by-step Solution
Detailed explanation
The \(n\)-th term of the given series is: \(T_n = \dfrac{1}{n} (1^2 + 2^2 + \ldots + n^2)\) Using the formula for the sum of squares of the first \(n\) natural numbers:…
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