**Linear Difference**

For a linear difference, the formula goes as such,

T=An + B

A=Difference.

'n'=the number of which you are solving

Tn= the number of 'n'

I'll explain it look:

n=..1, 2, 3, 5, 6

Tn=2, 3, 4, 5, 6

So here it would be, take any number from 'n'

If i take 'n' as 1, it would be

Tn= An + B

A=difference so since the difference between 2 or 3 is 1, 4 or 5 is 1, 5 or 6 is 1, then we will take 1.

If i take 'n' as 1, then the Tn of 1 is (2) so Tn = 2

2= (1)(1) + B

B=1

So you have the value of B now which is 1.

Now put it into the equation.

Tn= n +1

That is your formula.

**Quadratic Difference**

Okay, here's the one for quadratic equation.

This is a bit more complex.

An² + Bn + C= Tn

The principal is the same except for a few changes.

A= difference, but this time, the difference will be divided by two. It will ALWAYS be divided by two.

B & C are both unknowns, and therefore you will obtain two equations. You will have to solve them simultaneously. If you have any problems with solving the equation, then just private message me.

number: [1,] [ 2,] [3,] [4,] [5]

Tn=.......[4,] [7,] [12,] [19,] [28]

Okay this gets a bit complex. You have to find out the second difference.

So:

4, 7, 12, 19, 28

(3...5...7...9)

(....2...2...2)

What I have written in brackets are the differences between each 'Tn'

So you have a difference of 2 in the second difference.

In quadratic equations, you always divide the difference which is 'A' by 2.

So A=2/2, =1

First equation:

An² + Bn + C= Tn

I'm taking the first number now.

1(1)² + B(1) + C=4

1+B+C=4

(i)B+C=3 (Equation one)

An² + Bn + C= Tn

1(2)² + B(2) + C = 7

4 + 2B + C = 7

(ii)2B + C= 3 (Equation two)

Now, simultaneously

B=3-C (from equation one)

2(3-C) + C= 3

6- 2C + C = 3

6 -C = 3

C=3

B=0

Just solve it simultaneously.

Now put it into the equation

n²+3, there you go.