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

$v_n = v_{n-1} + 2n - 1$

$v_n = (v_{n-2} + 2n - 3) + 2n - 1 = v_{n-2} + 2\times2n - (1 + 3)$

$v_n = (v_{n-3} + 2n-5) + 2\times2n - (1 + 3) = v_{n-3} + 3\times2n - (1 + 3 + 5)$

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

$v_n = v_1 + (n-1)\times2n - (1 + 3 + 5 + \cdots + (2(n-1)-1))$

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

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

so that

$S + S' = \dfrac{(2n-2)(2n-1)}2$

and

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

So, we find

$S = (S + S') - S' = \dfrac{(2n-2)(2n-1)}2 - n(n-1) = (n-1)^2$

Then the n-th term to the sequence is

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