Verified answer

Answer:

To find the first 8 terms of the sequence with the general term \(A_n = n^2 + (n=1)\), you can plug in values of \(n\) from 1 to 8:

1. \(A_1 = 1^2 + 1 = 2\)

2. \(A_2 = 2^2 + 1 = 5\)

3. \(A_3 = 3^2 + 1 = 10\)

4. \(A_4 = 4^2 + 1 = 17\)

5. \(A_5 = 5^2 + 1 = 26\)

6. \(A_6 = 6^2 + 1 = 37\)

7. \(A_7 = 7^2 + 1 = 50\)

8. \(A_8 = 8^2 + 1 = 65\)

So, the first 8 terms of the sequence are: 2, 5, 10, 17, 26, 37, 50, 65.

[tex]\large\mathbb{SOLUTION :}[/tex]

To find the first 8 terms using the general term An = n² + (n = 1), we substitute the values of n from 1 to 8 into the equation.

For n = 1:

A1 = (1)² + (1 = 1)

A1 = 1 + 1

A1 = 2

For n = 2:

A2 = (2)² + (2 = 1)

A2 = 4 + 2

A2 = 6

For n = 3:

A3 = (3)² + (3 = 1)

A3 = 9 + 3

A3 = 12

For n = 4:

A4 = (4)² + (4 = 1)

A4 = 16 + 4

A4 = 20

For n = 5:

A5 = (5)² + (5 = 1)

A5 = 25 + 5

A5 = 30

For n = 6:

A6 = (6)² + (6 = 1)

A6 = 36 + 6

A6 = 42

For n = 7:

A7 = (7)² + (7 = 1)

A7 = 49 + 7

A7 = 56

For n = 8:

A8 = (8)² + (8 = 1)

A8 = 64 + 8

A8 = 72

Therefore, the first 8 terms using the general term An = n² + (n = 1) are:

2, 6, 12, 20, 30, 42, 56, 72.