Question

What is the value of \(a_{32}\) from the recurrence relation: \(a_1 = 4,\ a_n = 5n + a_{n-1}\ (n \ge 2)\)?

• A. 2369
• B. 2469
• C. 2569
• D. 2639

Hint:

Sum the recurrence from n = 2 to n = 32 using the arithmetic series formula, or derive a closed form for a_n.

Answer 1

The Correct Option is D

Step 1: The recurrence is \(a_n = a_{n-1} + 5n\) for \(n \ge 2\), with \(a_1 = 4\). Each term is obtained by adding \(5n\) to the previous term.

Step 2: Sum the recurrence from \(n = 2\) to \(n = 32\). Adding all these relations telescopes the left side:

\[ a_{32} - a_1 = \sum_{n=2}^{32} 5n = 5\sum_{n=2}^{32} n \]

Step 3: Compute \(\sum_{n=2}^{32} n\) using the arithmetic series formula for the numbers 2 through 32:

\[ \sum_{n=2}^{32} n = \frac{(2+32)\times 31}{2} = \frac{34 \times 31}{2} = 527 \]

Step 4: Substitute back to find \(a_{32}\):

\[ a_{32} = a_1 + 5 \times 527 = 4 + 2635 = 2639 \]

The value of \(a_{32}\) is 2639.

\[\boxed{a_{32} = 2639}\]