Sequence and Series
Special Series
JEE Main 2019
Grade 11
Question:
The sum $\frac{3 \times 1^3}{1^2} + \frac{5 \times (1^3 + 2^3)}{1^2 + 2^2} + \frac{7 \times (1^3 + 2^3 + 3^3)}{1^2 + 2^2 + 3^2} + \dots$ up to 9 terms is:
(1) $405$
(2) $495$
(3) $625$
(4) $900$
Step-by-Step Solution
Key Concept: Simplify the $n$-th term using $\sum k^3 = [\frac{n(n+1)}{2}]^2$ and $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$. The term reduces to $\frac{3n(n+1)}{2}$. Sum to $n = 9$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)