JEE Mains · Maths · STD 11 - 8. sequence and series
The sum \(\dfrac{1^3}{1} + \dfrac{1^3 + 2^3}{1 + 3} + \dfrac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \cdots\) up to 8 terms, is:
- A \(70\)
- B \(71\)
- C \(72\)
- D \(73\)
Answer & Solution
Correct Answer
(B) \(71\)
Step-by-step Solution
Detailed explanation
The \(n\)-th term of the given series is: \(T_n = \dfrac{1^3 + 2^3 + \cdots + n^3}{1 + 3 + 5 + \cdots + (2n - 1)}\) Using the formulas for the sum of cubes of first \(n\) natural numbers and the sum of first \(n\) odd numbers:…
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