Question

b) The first five terms in a different sequence are -29, -26, -21, -14, -5. Find, in terms of n, a formula for the nth term, V, of the sequence. ​

Answer 1

If
`v_n` is the n-th term in the sequence, observe that

`v_2 - v_1 = -26 - (-29) = 3`

`v_3 - v_2 = -21 - (-26) = 5`

`v_4 - v_3 = -14 - (-21) = 7`

`v_5 - v_4 = -5 - (-14) = 9`

and if the pattern continues,

`v_n - v_(n-1) = 2n - 1`

so the sequence is defined recursively by

`\begin{cases} v_1 = -29 \\ v_n = v_(n-1) + 2n - 1 & \text{for } n > 1 \end{cases}`

By this definition,

`v_(n-1) = v_(n-2) + 2(n-1) - 1 = v_(n-2) + 2n - 3`

`v_(n-2) = v_(n-3) + 2(n-2) - 1 = v_(n-3) + 2n - 5`

and so on. Then by substitution, we have

`v_n = v_(n-1) + 2n - 1`

`v_n = (v_(n-2) + 2n - 3) + 2n - 1 = v_(n-2) + 2*2n - (1 + 3)`

`v_n = (v_(n-3) + 2n-5) + 2*2n - (1 + 3) = v_(n-3) + 3*2n - (1 + 3 + 5)`

and if we keep doing this we'll eventually get
`v_n` in terms of
`v_1` to be

`v_n = v_1 + (n-1)*2n - (1 + 3 + 5 + \cdots + (2(n-1)-1))`

Evaluate the sum:

Let

`S = 1 + 3 + 5 + \cdots + (2(n-1)-1) = 1 + 3 + 5 + \cdots + (2n-3)`

`S' = 2 + 4 + 6 + \cdots + (2n - 2)`

Then

`S + S' = 1 + 2 + 3 + 4 + \cdots + (2n-3) + (2n-2) = \displaystyle \sum_(k=1)^(2n-2) k`

Recall that

`\displaystyle \sum_(i=1)^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2`

so that

`S + S' = \frac{(2n-2)(2n-1)}2`

and

`S' = 2 + 4 + 6 + \cdots + (2n-2) = 2 \left(1 + 2 + 3+ \cdots + (n-1)\right) = (n-1)n`

So, we find

`S = (S + S') - S' = \frac{(2n-2)(2n-1)}2 - n(n-1) = (n-1)^2`

Then the n-th term to the sequence is

`v_n = v_1 +2n(n-1) - S = -29 + 2n^2 - 2n - (n-1)^2 = \boxed{n^2-30}`