Difference Equations
Scott Morgan
April 1, 2022
1
Question
Create the general expression for a term in the sequence {−1, 2, 7, 14, ...}.
2
Answer
There are a number of ways to approach this problem. Here’s one.
Note that the difference between each pair of adjacent terms starts at 3 and
increases by two with each term.
If we let the nth term be sn could write the relationship between terms as
sn+1 = sn + 2n + 1
If we let s1 = −1, then s2 = 2, s3 = 7, etc.
Let’s rewrite the relationship between terms as sn+1 − sn = 2n + 1 and write
the differences for n, n-1, n-2, ..., all the way to 1 as follows.
So,
sn+1 − sn = 2n + 1
sn − sn−1 = 2(n − 1) + 1
sn−1 − sn−1 = 2(n − 2) + 1
...
s2 − s1 = 3
1
If we sum both sides of the equations, notice that one term on the left side
will cancel with a term in the row above. The sum on both sides will be:
sn+1 − s1 = [2n + 1] + [2(n − 1) + 1] + . . . + 3
Or,
sn+1 − s1 =
∞
X
2n + 1
n=1
Then,
sn+1 − s1 = 2
∞
X
n+
n=1
∞
X
1
n=1
n(n + 1)
+n
2
n(n + 1)
=2
+ n + s1
2
n(n + 1)
+ n + (−1)
=2
2
sn+1 − s1 = 2
sn+1
sn+1
sn+1 = n2 + 2n − 1
Doing a check we see that if we let n = 2, we get s2 = 2, as expected. Likewise,
s3 = 7, etc.
To get {sn } in terms of n, we substitute n − 1 for n.
s(n−1)+1 = (n − 1)2 + 2(n − 1) − 1
And,
sn = n2 − 2
Now, you can check that sn = {−1, 2, 7, 14, ...}.
2