sum of rth powers of the first n positive integers

sum of

r

th powers of the first

n

positive integers

A very classic problem is this: what is the sum of the numbers from

1

to

100

? The answer is

5050

, and there are a number of ways to see it. Before stating them, we generalize the problem somewhat: What is

∑

k

=

1

n

k

?

Theorem 1

.

∑

k

=

1

n

k

=

n

⁢

(

n

+

1

)

2

Proof.

Write the sum twice:

1

+

2

+

3

+

⋯

+

n

+

n

+

(

n

-

1

)

+

(

n

-

2

)

+

⋯

+

1

=

n

+

1

+

n

+

1

+

n

+

1

+

⋯

+

n

+

1

which is exactly

n

⁢

(

n

+

1

)

, so

∑

k

=

1

n

k

=

n

⁢

(

n

+

1

)

2

.

∎

Having seen this, it is natural to generalize this question further, and ask: What is

∑

k

=

1

n

k

r

, for any

integer

r

>

0

?

This question, being so general, does not have such a tidy answer, but it can be solved.

Theorem 2

.

Let

B

m

denote the

m

th

Bernoulli number

, and let

r

be a

positive integer

. Then

∑

k

=

0

n

k

r

=

∑

l

=

1

r

+

1

(

r

+

1

l

)

⁢

B

r

+

1

-

l

r

+

1

⁢

(

n

+

1

)

l

.

Observe that this formula is a

polynomial

in

n

of degree (

http://planetmath.org/OrderAndDegreeOfPolynomial

)

r

+

1

; for a given

r

, we can look up all the Bernoulli numbers we need and write down the polynomial explicitly. For example:

1

2

+

2

2

+

3

2

+

⋯

+

n

2

=

n

⁢

(

n

+

1

)

⁢

(

2

⁢

n

+

1

)

6

,

1

3

+

2

3

+

3

3

+

⋯

+

n

3

=

(

n

⁢

(

n

+

1

)

)

2

4

,

and

1

4

+

2

4

+

3

4

+

⋯

+

n

4

=

n

⁢

(

2

⁢

n

+

1

)

⁢

(

n

+

1

)

⁢

(

3

⁢

n

2

+

3

⁢

n

-

1

)

30

.

We will prove the result by a

generating function

argument

.

Proof.

For no obvious reason, let

f

n

⁢

(

x

)

:=

x

⁢

(

e

(

n

+

1

)

⁢

x

-

1

)

(

e

x

-

1

)

.

On the one hand, we have

f

n

⁢

(

x

)

=

x

⁢

1

-

(

e

x

)

n

+

1

1

-

e

x

=

x

⁢

∑

k

=

0

n

e

k

⁢

x

,

and so the coefficient of

x

r

+

1

is

∑

k

=

0

n

k

r

/

r

!

, more or less the sum we want.

On the other hand, using the definition of the Bernoulli numbers,

f

n

⁢

(

x

)

=

(

e

(

n

+

1

)

⁢

x

-

1

)

⁢

∑

j

=

0

∞

B

j

⁢

x

j

j

!

=

∑

l

=

1

∞

∑

j

=

0

∞

B

j

⁢

(

n

+

1

)

l

⁢

x

l

+

j

l

!

⁢

j

!

so the coefficient of

x

r

+

1

is

∑

l

=

1

r

+

1

(

n

+

1

)

l

l

!

⁢

B

r

+

1

-

l

(

r

+

1

-

l

)

!

.

Combining these two results, we get

∑

k

=

0

n

k

r

=

∑

l

=

1

r

+

1

(

r

+

1

l

)

⁢

B

r

+

1

-

l

r

+

1

⁢

(

n

+

1

)

l

.

∎

Observe that we need not use the Bernoulli numbers directly; any way we can extract the Taylor coefficients of

f

n

will give us the same results. In practical terms, we can hand

f

n

to a

computer algebra system

and it will print out the needed formula.

This proof is given as Exercise 2.23 in

http://www.math.upenn.edu/ wilf/DownldGF.html

generatingfunctionology.

Title

sum of

r

th powers of the first

n

positive integers

Canonical name

SumOfRthPowersOfTheFirstNPositiveIntegers

Date of creation

2013-03-22 14:14:38

Last modified on

2013-03-22 14:14:38

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

14

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 11B68

Classification

msc 05A15

Related topic

BernoulliNumber

Related topic

FormalPowerSeries

Related topic

ArithmeticProgression

Related topic

SumOfPowers

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