Question

1. (10 points) find the first five terms of the sequence, given the (n^{\text{th}}) term.

\\a_n = \frac{n - 4}{2n + 1}\\

1. (10 points) write the sum in sigma notation. you do not have to evaluate.

\\1(4) + 2(5) + 3(6) + \dots + 51(54) + 52(55)\\

Explanation:

Step1: Calculate the first term

$$a_1 = \frac{1 - 4}{2(1) + 1} = \frac{-3}{3} = -1$$

Step2: Calculate the second term

$$a_2 = \frac{2 - 4}{2(2) + 1} = -\frac{2}{5}$$

Step3: Calculate the third term

$$a_3 = \frac{3 - 4}{2(3) + 1} = -\frac{1}{7}$$

Step4: Calculate the fourth term

$$a_4 = \frac{4 - 4}{2(4) + 1} = 0$$

Step5: Calculate the fifth term

$$a_5 = \frac{5 - 4}{2(5) + 1} = \frac{1}{11}$$

Step6: Identify the general term pattern

$$i(i + 3)$$

Step7: Determine the summation limits

$$\sum_{i=1}^{52} i(i + 3)$$

Answer:

1. \(a_1 = -1\), \(a_2 = -\frac{2}{5}\), \(a_3 = -\frac{1}{7}\), \(a_4 = 0\), \(a_5 = \frac{1}{11}\)
2. \(\sum_{i=1}^{52} i(i + 3)\)