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 $=2times3^2+3times3=18+9=27$

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

This is correct

The 11th term would be

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

The 11th term $=275$

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

Method 1

Square numbers

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