Question

Consider the following:

Assertion (A): \(1^2 + 3^2 + 5^2 + \cdots + 39^2 = 10760\)

Reason (R): The sum of the squares of the first \(n\) odd numbers is \[ \frac{n(4n^2-1)}{3}. \]

The correct answer is:

• A. Both A and R are correct and R is the correct explanation of A
• B. Both A and R are correct and R is not correct explanation of A
• C. A is correct, but R is not correct
• D. A is incorrect, but R is correct

Hint:

To quickly sum squares of odd numbers up to a last term \( L \), use the alternative formula \( \frac{n(L)(L+2)}{3} \) where \( n \) is the number of terms and \( L \) is the last odd number. Here, \( \frac{20 \times 39 \times 41}{3} = 10660 \).

Answer 1

The Correct Option is D

Concept: The sum of a sequence can be calculated using the general term (\(T_k\)). For a sum of squares of odd numbers:
• General term \( T_k = (2k-1)^2 \).
• Sum formula: \( S_n = \sum_{k=1}^{n} T_k \).
• We use the standard summation formulas: \( \sum k^2 = \frac{n(n+1)(2n+1)}{6} \) and \( \sum k = \frac{n(n+1)}{2} \).

Step 1: Verifying Reason (R).
The sum of squares of the first \( n \) odd numbers is given by: \[ S_n = \sum_{k=1}^{n} (2k-1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) \] \[ S_n = 4 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 4 \left[ \frac{n(n+1)}{2} \right] + n \] Simplifying the expression: \[ S_n = \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} = \frac{n[2(2n^2+3n+1) - 6n - 6 + 3]}{3} = \frac{n(4n^2-1)}{3} \] Thus, Reason (R) is correct.

Step 2: Verifying Assertion (A).
The series is \( 1^2 + 3^2 + 5^2 + \dots + 39^2 \). The last term is \( 2n-1 = 39 \Rightarrow n = 20 \). Using the formula from R: \[ S_{20} = \frac{20(4(20)^2 - 1)}{3} = \frac{20(1600 - 1)}{3} = \frac{20 \times 1599}{3} \] \[ S_{20} = 20 \times 533 = 10660 \] Given value in Assertion is \( 10760 \). Since \( 10660 \neq 10760 \), Assertion (A) is incorrect.