Quadratic formula to find nth 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
and mathematicians also found that
Once you have found a, 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?
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?
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?
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?
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



