Question

1+(1+3)+(1+3+5)+(1+3+5+7)+... to 10 terms =

• A. 385
• B. 285
• C. 506
• D. 406

Hint:

Pattern recognition is very useful in series problems. The sum of the first \( n \) odd numbers equals \( n^2 \), while the sum of the first \( n \) even numbers equals \( n(n+1) \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

The series consists of terms where each term represents the sum of consecutive odd numbers. Our goal is to determine the general expression for the \( n \)-th term \( T_n \), and then compute the sum of these terms from \( n=1 \) to \( n=10 \).
Step 2: Key Formula or Approach:

1. The sum of the first \( n \) odd natural numbers equals \( n^2 \).
2. The formula for the sum of the first \( k \) squares is \( \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \).
Step 3: Detailed Explanation:

Observe the pattern in the terms of the sequence:

• 1st term (\( T_1 \)): \( 1 = 1^2 \)
• 2nd term (\( T_2 \)): \( 1+3 = 4 = 2^2 \)
• 3rd term (\( T_3 \)): \( 1+3+5 = 9 = 3^2 \)
• 4th term (\( T_4 \)): \( 1+3+5+7 = 16 = 4^2 \)

From this observation, the \( n \)-th term corresponds to the sum of the first \( n \) odd numbers, which is \[ T_n = n^2 \] Therefore, the required sum of the first 10 terms becomes \[ S_{10} = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} n^2 \] Applying the formula for the sum of squares: \[ \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \] Substituting \( k=10 \): \[ S_{10} = \frac{10(10+1)(2\cdot10+1)}{6} \] \[ S_{10} = \frac{10 \cdot 11 \cdot 21}{6} \] Simplifying step by step: \[ S_{10} = \frac{10}{2} \cdot 11 \cdot \frac{21}{3} \] \[ S_{10} = 5 \cdot 11 \cdot 7 \] \[ S_{10} = 55 \cdot 7 \] \[ S_{10} = 385 \]
Step 4: Final Answer:

The required sum is \( 385 \).