Mammoth Memory

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

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

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?

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=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?

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=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?

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=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?

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=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