# Quadratic formula to find `n^(th)` term of a sequence very easy method - consistent difference between differences

## Method 3

This method is very difficult to remember but very quick to work out.

**Example 1**

We know the following sequence is a quadratic sequence but what is the formula for the `n^(th)` term?

`a=(se\cond\ \ di\fference)/2=2/2=1`

Therefore `b=0`

and because `1^(st)` term `=a+b+c`

`1=1+0+c`

`1-1=0+c`

`c=0`

Summary `a=1` `b=0` `c=0`

therefore the formula for the `n^(th)` term in this sequence is

`an^2+bn+c`

or `1n^2+0n+0`

which is `n^2`

Now we need to check the formula is correct.

Try different values of `n` in the formula `n^2`

If `n=1` term `=1^2=1`

If `n=2` term `=2^2=4`

If `n=3` term `=3^2=9`

If `n=4` term `=4^2=16`

This is correct

Answer `=n^2`

**Example 2**

We know the following sequence is a quadratic sequence but what is the formula for the `n^(th)` term?

`a=(se\cond\ \ di\fference)/2=2/2=1`

Therefore `b=-1`

and because `1^(st)` term `=a+b+c`

`2=1-1+c`

`c=2`

Summary `a=1` `b=-1` `c=2`

therefore the formula for the `n^(th)` term in this sequence is

`an^2+bn+c`

or `1n^2+(-1)n+2`

which is `n^2-n+2`

Now we need to check the formula is correct.

Try different values of `n` in the formula `n^2-n+2`

If `n=1` term `=1^2-1+2=1-1+2=2`

If `n=2` term `=2^2-2+2=4-2+2=4`

If `n=3` term `=3^2-3+2=9-3+2=8`

If `n=4` term `=4^2-4+2=16-4+2=14`

This is correct

Answer `=n^2-n+2`

**Example 3**

We know the following sequence is a quadratic sequence but what is the formula for the `n^(th)` term?

`a=(se\cond\ \ di\fference)/2=2/2=1`

Therefore `b=0`

and because `1^(st)` term `=a+b+c`

`4=1+0+c`

`c=3`

Summarry `a=1` `b=0` `c=3`

therefore the formula for the `n^(th)` term in this sequence is

`an^2+bn+c`

or `1n^2+0n+3`

which is `n^2+3`

Now we need to check the formula is correct.

Try different values of `n` in the formula `n^2+3`

If `n=1` term `=1^2+3=1+3=4`

If `n=2` term `=2^2+3=4+3=7`

If `n=3` term `=3^2+3=9+3=12`

If `n=4` term `=4^2+3=16+3=19`

This is correct

Answer `=n^2+3`

**Example 4**

`a=(se\cond\ \ di\fference)/2=4/2=2`

Therefore `b=3`

and because `1^(st)` term `=a+b+c`

`5=2+3+c`

`c=5-2-3`

`c=0`

Summarry `a=2` `b=3` `c=0`

therefore the formula for the `n^(th)` term in this sequence is

`an^2+bn+c`

or `2n^2+3n+0`

which is `2n^2+3n`

Now we need to check the formula is correct.

Try different values of `n` in the formula `2n^2+3n`

If `n=1` term `=2times1^2+3times1=2+3=5`

If `n=2` term `=2times2^2+3times2=8+6=14`

If `n=3` term `=2times3^2+3times3=18+9=27`

If `n=4` term `=2times4^2+3times4=32+12=44`

This is correct

Answer `=2n^2+3n`

# Quadratic formula to find `n^(th)` term of a sequence very easy method - consistent difference between differences

## Method 3

This method is very difficult to remember but very quick to work out.

**Example 1**

`a=(se\cond\ \ di\fference)/2=2/2=1`

Therefore `b=0`

and because `1^(st)` term `=a+b+c`

`1=1+0+c`

`1-1=0+c`

`c=0`

Summary `a=1` `b=0` `c=0`

therefore the formula for the `n^(th)` term in this sequence is

`an^2+bn+c`

or `1n^2+0n+0`

which is `n^2`

Now we need to check the formula is correct.

Try different values of `n` in the formula `n^2`

If `n=1` term `=1^2=1`

If `n=2` term `=2^2=4`

If `n=3` term `=3^2=9`

If `n=4` term `=4^2=16`

This is correct

Answer `=n^2`

**Example 2**

`a=(se\cond\ \ di\fference)/2=2/2=1`

Therefore `b=-1`

and because `1^(st)` term `=a+b+c`

`2=1-1+c`

`c=2`

Summarry `a=1` `b=-1` `c=2`

therefore the formula for the `n^(th)` term in this sequence is

`an^2+bn+c`

or `1n^2+(-1)n+2`

which is `n^2-n+2`

Now we need to check the formula is correct.

Try different values of `n` in the formula `n^2-n+2`

If `n=1` term `=1^2-1+2=1-1+2=2`

If `n=2` term `=2^2-2+2=4-2+2=4`

If `n=3` term `=3^2-3+2=9-3+2=8`

If `n=4` term `=4^2-4+2=16-4+2=14`

This is correct

Answer `=n^2-n+2`

**Example 3**

`a=(se\cond\ \ di\fference)/2=2/2=1`

Therefore `b=0`

and because `1^(st)` term `=a+b+c`

`4=1+0+c`

`c=3`

Summarry `a=1` `b=0` `c=3`

therefore the formula for the `n^(th)` term in this sequence is

`an^2+bn+c`

or `1n^2+0n+3`

which is `n^2+3`

Now we need to check the formula is correct.

Try different values of `n` in the formula `n^2+3`

If `n=1` term `=1^2+3=1+3=4`

If `n=2` term `=2^2+3=4+3=7`

If `n=3` term `=3^2+3=9+3=12`

If `n=4` term `=4^2+3=16+3=19`

This is correct

Answer `=n^2+3`

**Example 4**

`a=(se\cond\ \ di\fference)/2=4/2=2`

Therefore `b=3`

and because `1^(st)` term `=a+b+c`

`5=2+3+c`

`c=5-2-3`

`c=0`

Summarry `a=2` `b=3` `c=0`

therefore the formula for the `n^(th)` term in this sequence is

`an^2+bn+c`

or `2n^2+3n+0`

which is `2n^2+3n`

Now we need to check the formula is correct.

Try different values of `n` in the formula `2n^2+3n`

If `n=1` term `=2times1^2+3times1=2+3=5`

If `n=2` term `=2times2^2+3times2=8+6=14`

If `n=3` term `=2times3^2+3times3=18+9=27`

If `n=4` term `=2times4^2+3times4=32+12=44`

This is correct

Answer `=2n^2+3n`