Question

The \( n^{th} \) term of the series \[ 1 + (3+5+7) + (9+11+13+15+17) + \dots \] is:

• A. \( (2n+1) \left[n^2 - (n-1)^2 \right] \)
• B. \( (2n-1) \left[(n-1)^2 - n^2 \right] \)
• C. \( (2n+1) \left[(n-1)^2 - n^2 \right] \)
• D. \( (2n-1) \left[(n-1)^2 + n^2 \right] \)

Hint:

Identify patterns in series and use standard summation formulas such as \( \sum k^2 \) and \( \sum k \) for simplifications.

Answer 1

The Correct Option is D

To determine the \( n^{th} \) term of the series: \[ 1 + (3+5+7) + (9+11+13+15+17) + \dots \] we first observe the pattern in the series:
1. The first group has 1 term: \( 1 \).
2. The second group has 3 terms: \( 3, 5, 7 \).
3. The third group has 5 terms: \( 9, 11, 13, 15, 17 \).
4. The fourth group would have 7 terms, and so on. Thus, the number of terms in the \( n^{th} \) group is \( 2n - 1 \). Next, we determine the starting number of the \( n^{th} \) group. The series consists of consecutive odd numbers. The total number of terms up to the \( (n-1)^{th} \) group is: \[ 1 + 3 + 5 + \dots + (2(n-1) - 1) = (n-1)^2 \] This is because the sum of the first \( k \) odd numbers is \( k^2 \). Therefore, the starting number of the \( n^{th} \) group is the next odd number after the last term of the \( (n-1)^{th} \) group. The last term of the \( (n-1)^{th} \) group is: \[ 2(n-1)^2 - 1 \] Thus, the starting number of the \( n^{th} \) group is: \[ 2(n-1)^2 + 1 \] The \( n^{th} \) group consists of \( 2n - 1 \) consecutive odd numbers starting from \( 2(n-1)^2 + 1 \). The sum of these \( 2n - 1 \) consecutive odd numbers is: \[ \text{Sum} = (2n - 1) \left[2(n-1)^2 + 1 + 2(2n - 1 - 1)\right] / 2 \] However, a simpler approach is to recognize that the sum of \( 2n - 1 \) consecutive odd numbers starting from \( a \) is: \[ \text{Sum} = (2n - 1) \left[a + (2n - 2)\right] \] Substituting \( a = 2(n-1)^2 + 1 \): \[ \text{Sum} = (2n - 1) \left[2(n-1)^2 + 1 + (2n - 2)\right] \] Simplifying the expression inside the brackets: \[ 2(n-1)^2 + 1 + 2n - 2 = 2(n-1)^2 + 2n - 1 \] Thus, the sum of the \( n^{th} \) group is: \[ (2n - 1) \left[2(n-1)^2 + 2n - 1\right] \] However, upon closer inspection, we realize that the sum of the \( n^{th} \) group can be expressed more simply as: \[ (2n - 1) \left[(n-1)^2 + n^2\right] \] This matches option (4). Therefore, the \( n^{th} \) term of the series is: \[ \boxed{(2n-1) \left[(n-1)^2 + n^2 \right]} \]