Question

(iv) $t_n=n^2-2n$

Answer 1

(iv)
We have the n^th term of a sequence as t_n = n² – 2n.
To find the first five terms of the sequence, we need to take n = 1, 2, 3, 4 and 5.

For n = 1, we have:
t₁ = 1² – 2 × 1 = 1 – 2 = –1
For n = 2, we have:
t₂ = 2² – 2 × 2 = 4 – 4 = 0
For n = 3, we have:
t₃ = 3² – 2 × 3 = 9 – 6 = 3
For n = 4, we have:
t₄ = 4² – 2 × 4 = 16 – 8 = 8
For n = 5, we have:
t₅ = 5² – 2 × 5 = 25 – 10 = 15

Thus, the first five terms of the sequence are –1, 0, 3, 8 and 15.