Question

Sum of the series \( 1(1) + 2(1+3) + 3(1+3+5) + 4(1+3+5+7) + \cdots + 10(1+3+5+7+\cdots+19) \) is equal to:

• A. \( 385 \)
• B. \( 1025 \)
• C. \( 1125 \)
• D. \( 2025 \)
• E. \( 3025 \)

Hint:

To quickly square a number ending in 5 (like 55): 1. Multiply the first digit (5) by the next integer (6) to get 30. 2. Append 25 at the end. Result: 3025.

Answer 1

Answer: (E)

Concept: The sum of the first \( n \) odd natural numbers is always a perfect square: \[ 1 + 3 + 5 + \cdots + (2n-1) = n^2 \] The general term \( T_r \) of the given series can be written as \( r \times (\text{sum of first } r \text{ odd numbers}) \).

Step 1: Find the general term \( T_r \).
The \( r^{th} \) term is: \[ T_r = r \times (r^2) = r^3 \]

Step 2: Calculate the total sum for \( n=10 \).
The total sum \( S \) is the summation of \( T_r \) from \( r=1 \) to \( 10 \): \[ S = \sum_{r=1}^{10} r^3 \] The formula for the sum of cubes of first \( n \) natural numbers is: \[ \sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} \right]^2 \]

Step 3: Plug in the value \( n=10 \).
\[ S = \left[ \frac{10(11)}{2} \right]^2 \] \[ S = (5 \times 11)^2 = (55)^2 \] \[ S = 3025 \]