, write it out in a line with gaps between each term:

2

00000

6

00000

12

00000

20

Add a line of term below listing the differences between each pair of terms:

2

00000

6

00000

12

00000

20

000

4

00000

6

000000

8

Add another line of term below listing the differences between each pair of terms:

2

00000

6

00000

12

00000

20

000

4

00000

6

000000

8

000000

2

000000

2

Notice that both terms of the last line are the same.

That implies that our given sequence can be matched using a quadratic formula, which we can construct using the first term of each of the lines as a coefficient:

a

n

=

2

0

!

+

4

1

!

(

n

−

1

)

+

2

2

!

(

n

−

1

)

(

n

−

2

)

a

n

=

2

+

4

n

−

4

+

n

2

−

3

n

+

2

a

n

=

n

2

+

n

If you know the formulas for

N

∑

n

=

1

n

2

and

N

∑

n

=

1

n

then you can just add them to provide the formula for

N

∑

n

=

1

a

n

.

I would rather construct it directly:

Add another line to the top of our sequences, consisting of the sums of the terms on the original top line:

0

00000

2

00000

8

00000

20

00000

40

000

2

00000

6

00000

12

00000

20

000000

4

00000

6

000000

8

000000000

2

000000

2

We want to start with

s

1

=

2

. So ignore the first terms of each sequence and use the second terms to give us our formula for the sum to

## Answers ( )

Given the sequence

2

,

6

,

12

,

20

, write it out in a line with gaps between each term:

2

00000

6

00000

12

00000

20

Add a line of term below listing the differences between each pair of terms:

2

00000

6

00000

12

00000

20

000

4

00000

6

000000

8

Add another line of term below listing the differences between each pair of terms:

2

00000

6

00000

12

00000

20

000

4

00000

6

000000

8

000000

2

000000

2

Notice that both terms of the last line are the same.

That implies that our given sequence can be matched using a quadratic formula, which we can construct using the first term of each of the lines as a coefficient:

a

n

=

2

0

!

+

4

1

!

(

n

−

1

)

+

2

2

!

(

n

−

1

)

(

n

−

2

)

a

n

=

2

+

4

n

−

4

+

n

2

−

3

n

+

2

a

n

=

n

2

+

n

If you know the formulas for

N

∑

n

=

1

n

2

and

N

∑

n

=

1

n

then you can just add them to provide the formula for

N

∑

n

=

1

a

n

.

I would rather construct it directly:

Add another line to the top of our sequences, consisting of the sums of the terms on the original top line:

0

00000

2

00000

8

00000

20

00000

40

000

2

00000

6

00000

12

00000

20

000000

4

00000

6

000000

8

000000000

2

000000

2

We want to start with

s

1

=

2

. So ignore the first terms of each sequence and use the second terms to give us our formula for the sum to

n

terms:

s

n

=

2

0

!

+

6

1

!

(

n

−

1

)

+

6

2

!

(

n

−

1

)

(

n

−

2

)

+

2

3

!

(

n

−

1

)

(

n

−

2

)

(

n

−

3

)

s

n

=

2

+

6

n

−

6

+

3

n

2

−

9

n

+

6

+

1

3

n

3

−

2

n

2

+

11

3

n

−

2

s

n

=

1

3

(

n

3

+

3

n

2

+

2

n

)

s

n

=

1

3

n

(

n

+

1

)

(

n

+

2

)