Nth term sequece 5 6 8 11 15 and show working?

Answer 1

t(n) = 5 6 8 11 15

First differences = 1 2 3 4

Second differences = 1 1 1

So the sequence is quadratic with the coefficient of n2 being half of the second difference ie half of 1.

So t(1) = 0.5*12 + b*1 + c = 5 ie b + c = 4.5

and

So t(2) = 0.5*22 + b*2 + c = 6 ie 2b + c = 4

Solving these two equations simultaneously, b = -0.5 and c = 5

Therefore, the sequence is t(n) = (n2 - n + 10)/2