n T ∆1 ∆2
1 -1
6
2 5 2
8
3 13 2
10
4 23 2
12
5 35 2
14
6 49
In the table above the column n is the position of the term T, in the next column, in the series. ∆1 is the first difference—the difference between consecutive terms. ∆2 is the second difference—the difference between consecutive ∆1s, which we can see is a constant 2. That means we found a pattern. It tells us that we can find a polynomial of degree 2 (because it’s the 2nd difference ∆2 that is constant). Such a polynomial is a quadratic in n:
T(n)=an²+bn+c where a, b, c are constants to be found.
From the table we can use any three corresponding values of T and n to find a, b and c.
Let’s use n=1, 2, 3 and T(1), T(2) and T(3):
n=1: T(1)=a+b+c=-1
n=2: T(2)=4a+2b+c=5
n=3: T(3)=9a+3b+c=13
T(2)-T(1): 3a+b=6
T(3)-T(1): 8a+2b=14, or, dividing by 2, 4a+b=7.
Subtract 3a+b=6 from this and we get a=1.
So b=6-3a=6-3=3.
And finally, c=-1-a-b=-1-1-3=-5.
Therefore T(n)=n²+3n-5.
As a test put n=6:
T(6)=36+18-5=49, which matches the given value.