Quadratic formula to find `n^(th)` term of a sequence easier method - consistent difference between difference

Method 2

This method is trickier to remember but less work to find an answer.

We know

Taking the 1st and 2nd steps form the summary, this method is trickier but is more quicker to an answer

and mathematicians also found that

Mathematicians also found that a is the second difference divided by 2

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

`an^2+bn+c`

Example 1

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

Find the nth term to this equation using method 2

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

so `1=a+b+c`

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

so `4=4a+2b+c`

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

`a=2/2=1`

Now we have

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

`4=4a+2b+c` ................ (2)

and `a=1`

therefore equation (1) becomes `1=1+b+c`

`1-1=b+c`

`-b=c`

therefore equation (2) becomes `4=4times1+2b-b`

`4-4=2b-b`

`0=2b-b`

`0=1b`

`b=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

Answer `=n^2`

Example 2

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

Find the nth term to this equation using method 2

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

so `2=a+b+c`

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

so `4=4a+2b+c`

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

`a=2/2=1`

Now we have

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

`4=4a+2b+c` ................ (2)

and `a=1`

therefore equation (1) becomes `2=1+b+c`

`1=b+c`........... (i)

therefore equation (2) becomes `4=4times1+2b+c`

`4=4+2b+c`

`4-4=2b+c`

`0=2b+c` ......... (ii)

Now we have two simultaneous equations:

`1=b+c` ........... (i)

`0=2b+c` ......... (ii)

From (i) we get `b=1-c`

Substitute that into (ii) we get

`0=2(1-c)+c`

`0=2-2c+c`

`0=2-1c`

`-2=-c`

`c=2`

Now substitute `c=2` into (i)

`1=b+2`

`b=1-2`

`b=-1`

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

Example 3

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

Find the nth term to this equation using method 2

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

so `4=a+b+c`

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

so `7=4a+2b+c`

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

`a=2/2=1`

Now we have

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

`7=4a+2b+c` .............. (2)

and `a=1`

therefore equation (1) becomes `4=1+b+c`

`3=b+c`........... (i)

therefore equation (2) becomes `7=4times1+2b+c`

`7=4+2b+c`

`7-4=2b+c`

`3=2b+c` ......... (ii)

Now we have two simultaneous equations:

`3=b+c` ........... (i)

`3=2b+c` ......... (ii)

From (i) we get `b=3-c`

Substitute that into (ii) we get

`3=2(3-c)+c`

`3=6-2c+c`

`3-6=-2c+c`

`-3=-c`

`c=-3/-1=3`

Now substitute `c=3` into (i)

`3=b+3`

`b=3-3`

`b=0`

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=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 using an easier method?

Find the nth term to this equation using method 2

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

so `5=a+b+c`

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

so `14=4a+2b+c`

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

`a=4/2=2`

Now we have

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

`14=4a+2b+c` ................ (2)

and `a=2`

therefore equation (1) becomes `5=2+b+c`

`3=b+c`........... (i)

therefore equation (2) becomes `14=4times2+2b+c`

`14=8+2b+c`

`6=2b+c` ......... (ii)

Now we have two simultaneous equations:

`3=b+c` ........... (i)

`6=2b+c` ......... (ii)

From (i) we get `b=3-c`

Substitute that into (ii) we get

`6=2(3-c)+c`

`6=6-2c+c`

`6-6=-2c+c`

`0=-c`

`c=0`

Now substitute `c=0` into (i)

`3=b+c`

`3=b+0`

`b=3`

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+3times3=18+9=27`

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

This is correct

Answer `2n^2+3n`