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.

We want to find `a n^2 + bn  +  c`

Note: When we say second level difference we mean second level difference between first term and second term on the second level. 

Note: When we say first first level difference we mean first level difference between the first term and second on the first row.

Summary

`a = (2\text(D))/2`       `b = 1\text(D) - 3a`       `c =` 1^st term `- a - b`

D = difference between 1^st and 2^nd term on 1^st or ^2nd level.

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=(5-3)/2=2/2 = 1`

`b  =` first level difference `- 3a`

`= (4-1) - 3`  x  `1 = 3 - 3 = 0`

`b = 0`

`c =` 1^st term `-a - b = 1 - 1 - 0 = 0` 

`c = 0`

`an^2 + bn + c`

`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= (4-2)/2 = 2/2=1`

`b =` first level difference `- 3a`

`b = (4 - 2) - 3`  x  `1 =`

`b = 2 - 3 = -1`

`c =` 1^st term `- a - b = 2 - 1 - (-1) = 2`

`an^2 + bn + c`

`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=(5 - 3)/2 = 2/2 =1`

`b =` first level difference `- 3a`

`b = (7 - 4) - 3`  x  `1`

`b = 3 - 3 = 0`

`c =` 1^st term `- a - b = 4 - 1 - 0 = 3`

`an^2 + bn + c`

`n^2 - 0 + 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

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= (13 - 9)/2 = 4/2 = 2`

`b =` first level difference `- 3a`

`= (14 - 5) - 3`  x  `2 = 9 - 6 - 3`

`b = 3`

`c =` 1^st term `- a - b = 5 - 2 - 3 = 0`

`an^2 + bn + c`

`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`