Mammoth Memory

Using the quadratic formula to find `n^(th)` term of a sequence - consistent difference between differences

 

Method 1

This is the easiest to remember but takes the most time to work out an answer.

We know (from the last section) that:

Summary of 1st, 2nd and 3rd quadratic terms

Once you have found `a,b,\ &\ c` you can slot these figures in the following formula.

`an^2+bn+c` 

Example 1

Square numbers

We know the following sequence is a quadratic sequence but what is the formula for the `n^(th)`  term and what is the `14^(th)` term?

What is the 14th term of this sequence using method 1 

The 1st term `=a+b+c`

so                 `1=a+b+c`

 

The 2nd term `=4a+2b+c`

so                   `4=4a+2b+c`

 

The 3rd term `=9a+3b+c`

so                   `9=9a+3b+c`

 

Now you have 3 simultaneous equations.

`1=a+b+c`  ....................  (1)

`4=4a+2b+c`  ................  (2)

`9=9a+3b+c`  ................  (3)

Now solve for one letter at a time

`1=a+b+c`

therefore     `c=1-a-b`

Substitute this into equations (2) and (3).

`4=4a+2b+1-a-b`

`9=9a+3b+1-a-b`

These two simultaneous equations become

`4-1=4a-a+2b-b`

and               `9-1=9a-a+3b-b`

Which simplify to

  `3=3a+b`   ........... (i)

and  `8=8a+2b`   .......... (ii)

 

Now we can solve these two simultaneous equations

    `3=3a+b`

is the same as `b=3-3a`

substitute this into (ii)

  `8=8a+2(3-3a)`

which is         `8=8a+6-6a`

which is `8-6=8a-6a`

becomes       `2=2a`

therefore      `a=1`

 

Sustitute `a=1` into (i)

and          `3=3a+b`

becomes `3=3times1+b`

   `b=3-3=0`

 

Now substitute `a=1` and `b=0` into equation (1)

`1=a+b+c`

`1=1+0+c`

`c=1-1=0`

Summary `a=1`   `b=0`   and   `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

The 14th term would be

If  `n=14`       then  `n^2=14^2=196`

The 14th term `=196`

 

 

Example 2

We know the following sequence is a quadratic sequence but what is the formula of the `n^(th)` term and what is the 16th term?

What is the 16th term of this sequence using method 1 

The 1st term `=a+b+c`

so                 `2=a+b+c`

 

The 2nd term `=4a+2b+c`

so                   `4=4a+2b+c`

 

The 3rd term `=9a+3b+c`

so                  `8=9a+3b+c`

 

Now you have 3 simultaneous equations.

`2=a+b+c`  ....................  (1)

`4=4a+2b+c`  ................  (2)

`8=9a+3b+c`  ................  (3)

Now solve for one letter at a time

`2=a+b+c`

therefore     `c=2-a-b`

Substitute this into equations (2) and (3).

`4=4a+2b+(2-a-b)`

`8=9a+3b+(2-a-b)`

These two simultaneous equations become

`4-2=4a-a+2b-b`

and               `8-2=9a-a+3b-b`

Which simplify to

  `2=3a+b`   ............ (i)

and  `6=8a+2b`   .......... (ii)

 

Now we can solve these two simultaneous equations

    `2=3a+b`

is the same as `b=2-3a`

substitute this into (ii)

  `6=8a+2(2-3a)`

which is         `6=8a+4-6a`

which is `6-4=8a-6a`

becomes       `2=2a`

therefore      `a=1`

 

Sustitute `a=1` into (i)

and          `2=3a+b`

becomes `2=3times1+b`

   `b=2-3=-1`

 

Now substitute `a=1` and `b=-1` into equation 1

`2=a+b+c`

`2=1-1+c`

`c=2`

Summary `a=1`   `b=-1`   and   `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=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

The 16th term would be

If  `n=16`       then  `n^2-n+2=16^2-16+2=256-16+2=242`

The 16th term `=242`

 

 

Example 3

We know the following sequence is a quadratic sequence but what is the formula of the `n^(th)` term and what is the 12th term?

What is the 12th term of this sequence using method 1 

The 1st term `=a+b+c`

so                  `4=a+b+c`

 

The 2nd term `=4a+2b+c`

so                   `7=4a+2b+c`

 

The 3rd term `=9a+3b+c`

so                `12=9a+3b+c`

 

Now you have 3 simultaneous equations.

  `4=a+b+c`  .....................  (1)

  `7=4a+2b+c`  ................  (2)

`12=9a+3b+c`  ................  (3)

Now solve for one letter at a time

`4=a+b+c`

therefore     `c=4-a-b`

Substitute this into equations (2) and (3).

  `7=4a+2b+(4-a-b)`

`12=9a+3b+(4-a-b)`

These two simultaneous equations become

  `7-4=4a-a+2b-b`

and               `12-4=9a-a+3b-b`

Which simplify to

  `3=3a+b`   ........... (i)

and  `8=8a+2b`   .......... (ii)

 

Now we can solve these two simultaneous equations

    `3=3a+b`

is the same as `b=3-3a`

substitute this into (ii)

  `8=8a+2(3-3a)`

which is         `8=8a+6-6a`

which is `8-6=8a-6a`

becomes       `2=2a`

therefore      `a=2/2=1`

 

Sustitute `a=1` into (i)

and          `3=3a+b`

becomes `3=3times1+b`

   `b=3-3=0`

 

Now substitute `a=1` and `b=0` into equation (1)

`4=a+b+c`

`4=1+0+c`

`c=4-1`

`c=3`

Summary `a=1`   `b=0`   and   `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

The 12th term would be

If  `n=12`       then  `n^2+3=12^2+3`

`=144+3`

`=147`

The 12th term `=147`

 

 

Example 4

We know the following sequence is a quadratic sequence but what is the formula of the `n^(th)` term and what is the 11th term?

What is the 11th term of this sequence using method 1 

The 1st term `=a+b+c`

so                 `5=a+b+c`

 

The 2nd term `=4a+2b+c`

so                 `14=4a+2b+c`

 

The 3rd term `=9a+3b+c`

so                `27=9a+3b+c`

 

Now you have 3 simultaneous equations.

  `5=a+b+c`  ....................   (1)

`14=4a+2b+c`  ................  (2)

`27=9a+3b+c`  ................  (3)

Now solve for one letter at a time

`5=a+b+c`

therefore     `c=5-a-b`

Substitute this into equations (2) and (3).

`14=4a+2b+5-a-b`

`27=9a+3b+5-a-b`

These two simultaneous equations become

`14-5=4a-a+2b-b`

and               `27-5=9a-a+3b-b`

Which simplify to

    `9=3a+b`   ............ (i)

and  `22=8a+2b`   .......... (ii)

 

Now we can solve these two simultaneous equations

    `9=3a+b`

is the same as `b=9-3a`

substitute this into (ii)

     `22=8a+2(9-3a)`

which is            `22=8a+18-6a`

which is `22-18=8a-6a`

becomes            `4=2a`

therefore           `a=4/2=2`

 

Sustitute `a=2` into (i)

and          `9=3a+b`

becomes `9=3times2+b`

   `b=9-6=3`

 

Now substitute `a=2` and `b=3` into equation (1)

`5=a+b+c`

`5=2+3+c`

`c=0`

Summary `a=2`   `b=3`   and   `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+3times2=18+9=27`

If  `n=4`       term  `=2times4^2+3times2=32+12=44`

This is correct

The 11th term would be

If  `n=11`       then  `2n^2+3n=2times11^2+3times11=242+33=275`

The 11th term `=275`

More Info