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, b, & c you can slot these figures in the following formula
an2+bn+c
Example 1
We know the following sequence is a quadratic sequence but what is the formula for the nth 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=second difference2
a=22=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=4×1+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 nth term in this sequence is
an2+bn+c
or 1n2+0n+0
which is n2
Now we need to check the formula is correct.
Try different values of n in the formula n2
If n=1 term =12=1
If n=2 term =22=4
If n=3 term =32=9
If n=4 term =42=16
This is correct
Answer =n2
Example 2
We know the following sequence is a quadratic sequence but what is the formula for the nth 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=second difference2
a=22=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=4×1+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 nth term in this sequence is
an2+bn+c
or 1n2+(-1)n+2
which is n2-n+2
Now we need to check the formula is correct.
Try different values of n in the formula n2-n+2
If n=1 term =12-1+2=2
If n=2 term =22-2+2=4-2+2=4
If n=3 term =32-3+2=9-3+2=8
If n=4 term =42-4+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 nth 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=second difference2
a=22=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=4×1+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 nth term in this sequence is
an2+bn+c
or 1n2+0n+3
which is n2+3
Now we need to check the formula is correct.
Try different values of n in the formula n2+3
If n=1 term =12+3=4
If n=2 term =22+3=4+3=7
If n=3 term =32+3=9+3=12
If n=4 term =42+3=16+3=19
This is correct
Answer n2+3
Example 4
We know the following sequence is a quadratic sequence but what is the formula for the nth 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=second difference2
a=42=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=4×2+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 nth term in this sequence is
an2+bn+c
or 2n2+3n+0
which is 2n2+3n
Now we need to check the formula is correct.
Try different values of n in the formula 2n2+3n
If n=1 term =2×12+3×1=2+3=5
If n=2 term =2×22+3×2=8+6=14
If n=3 term =2×32+3×3=18+9=27
If n=4 term =2×42+3×4=32+12=44
This is correct
Answer 2n2+3n



