Question

Find the nth term of the sequence 7,25,51,85,127​

Answer 1

Let a (n) denote the n-th term of the given sequence.

Check the forward differences, and denote the n-th difference by b (n). That is,

b (n) = a (n + 1) - a (n)

These so-called first differences are

b (1) = a (2) - a (1) = 25 - 7 = 18

b (2) = a (3) - a (2) = 51 - 25 = 26

b (3) = a (4) - a (3) = 85 - 51 = 34

b (4) = a (5) - a (4) = 127 - 85 = 42

Now consider this sequence of differences,

18, 26, 34, 42, …

and notice that the difference between consecutive terms in this sequence b is 8:

26 - 18 = 8

34 - 26 = 8

42 - 34 = 8

and so on. This means b is an arithmetic sequence, and in particular follows the rule

b (n) = 18 + 8 (n - 1) = 8n + 10

for n ≥ 1.

So we have

a (n + 1) - a (n) = 8n + 10

or, replacing n + 1 with n,

a (n) = a (n - 1) + 8 (n - 1) + 10

a (n) = a (n - 1) + 8n + 2

We can solve for a (n) by iteratively substituting:

a (n) = [a (n - 2) + 8 (n - 1) + 2] + 8n + 2

a (n) = a (n - 2) + 8 (n + (n - 1)) + 2×2

a (n) = [a (n - 3) + 8 (n - 2) + 2] + 8 (n + (n - 1)) + 2×2

a (n) = a (n - 3) + 8 (n + (n - 1) + (n - 2)) + 3×2

and so on. The pattern should be clear; we end up with

a (n) = a (1) + 8 (n + (n - 1) + … + 3 + 2) + (n - 1)×2

The middle group is the sum,

$\displaystyle 8\sum_{k=2}^nk=8\sum_{k=1}^nk-8=\frac{8n(n+1)}2-8=4n^2+4n-8$

so that

a (n) = a (1) + (4n ² + 4n - 8) + 2 (n - 1)

a (n) = 4n ² + 6n - 3