1/2n² + 1/2n + 1

This is a hard one and maybe closer to a maths challenge question. It is easier to explain in person.

First common difference is 1 then 2 then 3 and so on. The second common difference is 1. As there is a constant second difference we know the nth term follows the equation an² + bn + c.

We can then sub 1 into this equation so ax1² + bx1 c = 2

Therefore : a + b + c = 2 Sub 2 and 3 in to get three equations with three unknowns : 4a + 2b + c = 4 AND 9a + 3b + c = 7

Get everything in terms of c so c = 2 - a - b = 4 - 4a - 2b = 7 - 9a - 3b.

Comparing two of these equations to each other eliminates c as an unknown for a while eg 2 - a - b = 4 - 4a - 2b becomes b+3a=2

Compare two other equations to get another expression of a and b so : 7 - 9a - 3b = 4 - 4a - 2b becomes 3=5a+b

We now have 5a+b=3 and 3a+b=2. We can use a simultaneous equation to minus the latter from the former and end with 2a=1.

Therefore a =1/2 Sub this into 3a+b=2 and we get b = 1/2

if a and b are a half and a+b+c=2 then c=1.

Therefore answer is 1/2n² + 1/2n + 1

(2x+3)^2

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