Question

Find the sum of the sequence: \( 2 + 3 + 5 + 6 + 8 + 9 + \dots + 2n \) terms.

• A. \( 3n^2 + 2n \)
• B. \( 4n^2 + 2n \)
• C. \( 4n^2 \)
• D. \( 5n^2 + 2n \)

Hint:

To sum sequences with alternating patterns, split them into separate arithmetic progressions and sum each individually.

Answer 1

The Correct Option is A

Step 1: Identifying the Pattern The sequence consists of alternating even and odd terms: \[ 2, 3, 5, 6, 8, 9, \dots \] Grouping them: - The even-indexed terms form: \( 2, 5, 8, \dots \) - The odd-indexed terms form: \( 3, 6, 9, \dots \)
Step 2: Sum of Even-Indexed Terms This sequence follows an arithmetic progression with: \[ a = 2, \quad d = 3 \] The sum of the first \( n \) terms of an AP is: \[ S_n = \frac{n}{2} \times (2a + (n-1)d) \] Substituting values: \[ S_{\text{even}} = \frac{n}{2} \times (2(2) + (n-1)(3)) = \frac{n}{2} \times (4 + 3n - 3) = \frac{n}{2} \times (3n + 1) = \frac{3n^2 + n}{2} \]
Step 3: Sum of Odd-Indexed Terms For the sequence \( 3, 6, 9, \dots \), with \( a = 3 \) and \( d = 3 \), using the same sum formula: \[ S_{\text{odd}} = \frac{n}{2} \times (2(3) + (n-1)(3)) = \frac{n}{2} \times (6 + 3n - 3) = \frac{n}{2} \times (3n + 3) = \frac{3n^2 + 3n}{2} \]
Step 4: Total Sum of the Sequence Adding both: \[ S = S_{\text{even}} + S_{\text{odd}} = \frac{3n^2 + n}{2} + \frac{3n^2 + 3n}{2} \] \[ S = \frac{6n^2 + 4n}{2} = 3n^2 + 2n \] Thus, the required sum is: \[ \boxed{3n^2 + 2n} \]

Answer 2

The sequence provided is \( 2 + 3 + 5 + 6 + 8 + 9 + \dots + 2n \) terms. We need to determine the sum of this sequence. Observing the sequence, it alternates between +1 and +2 starting from 2, meaning it can be split into two subsequences:

• Even-indexed terms: \(2, 5, 8, \dots\)
• Odd-indexed terms: \(3, 6, 9, \dots\)

The ith term in the even-indexed subsequence follows the formula \(t_i = 2 + 3(i-1)\). For the odd-indexed terms, it follows \(t_i = 3 + 3(i-1)\).

Given that there are \(2n\) terms in total, each subsequence will have n terms. The sum of an arithmetic sequence is calculated as follows:

Sum of even-indexed terms:

\(S_e = \frac{n}{2} \times \text{(first term + last term)}\)

\(S_e = \frac{n}{2} \times \left[2 + (2 + 3(n-1))\right] = \frac{n}{2} \times (2 + 3n - 1) = \frac{n}{2} \times (3n + 1)\)

\(S_e = \frac{3n^2+n}{2}\)

Sum of odd-indexed terms:

\(S_o = \frac{n}{2} \times \text{(first term + last term)}\)

\(S_o = \frac{n}{2} \times \left[3 + (3 + 3(n-1))\right] = \frac{n}{2} \times (3 + 3n - 1) = \frac{n}{2} \times (3n + 2)\)

\(S_o = \frac{3n^2+2n}{2}\)

Total sum:

\(S = S_e + S_o\)

\(S = \frac{3n^2+n}{2} + \frac{3n^2+2n}{2} = \frac{6n^2+3n}{2}\)

\(S = 3n^2 + \frac{3n}{2} + \frac{2n}{2}\)

\(S = 3n^2 + 2n\)

This matches the correct answer given: \(3n^2 + 2n\).