Sum of Consecutive Powers (3rd, 4th, 5th, and 6th) Calculator & Formulae



Consecutive Power Sum Calculator
Find the sum of consecutive

3rd powers       4th powers
5th powers       6th powers

from    to     inclusive.

The sum of consecutive cubes, 4th powers, 5th powers, 6th powers, etc. are known as power sums. The power sum from 1 to n with power p (1p + 2p + 3p + ... + np) is denoted by Sp(n).

For positive integer values of p and n, the value of Sp(n) can be found by evaluating a polynomial in n. Once you know how to sum the powers from 1 to n, you can find the sum of powers between any two values a and b. These equations are explained below, and you can also use the power sum calculator on the left.

Power Sum Equations

The sum of consecutive cubes from 1³ to n³ is the square of the nth triangular number. That is,

S₃(n) = n²(n+1)²/4 = [n(n+1)/2]² = [T(n)]²

The sum of 4th powers is

S₄(n) = n(n+1)(2n+1)(3n² + 3n - 1)/30

The sum of 5th powers is

S₅(n) = n²(n+1)²(2n² + 2n - 1)/12

The sum of 6th powers is

S₆(n) = n(n+1)(2n+1)(3n⁴ + 6n³ - 3n + 1)/42

Sum of Consecutive Powers from ap to bp

The sum of any sequence of consecutive powers from ap to bp (inclusive) is the difference between Sp(b) and Sp(a) plus the ap (You need to add ap because the sum is inclusive of the endpoints ap and bp .)

Example: Find the sum of all the 5th powers between 100⁵ and 120⁵, including the endpoints.

S₅(120) - S₅(100) + 100⁵
= 120²(121²)(29039)/12 - 100²(101²)(20199)/12 + 10000000000
= 510191998800 - 171708332500 + 10000000000
= 348,483,666,300

Fun Facts About Power Sums

0. S₀(n) is the sum of the 0th powers from 1⁰ to n⁰. Since any non-zero number raised to the 0th power equals 1, S₀(n) = (1)(n) = n.


1. If you know the polynomials S₀(n), S₁(n), S₂(n),..., Sp-1(n), then you can find Sp(n) by expanding the sum Σj=1...n(1+j)p+1 into individual power sums.

Σj=1...n (1+j)p+1 = Σk=0...p+1[Σj=1...n C(p+1,k)jk] = Σk=0...p+1 C(p+1,k)Sk(n)

where C(p+1,k) is the binomial coefficient p+1 choose k. If we continue to simplify the expression, we get

Σj=1...n (1+j)p+1 - Sp+1(n) = Σk=0...p C(p+1,k)Sk(n)

(n+1)p+1 - 1 = Σk=0...p C(p+1,k)Sk(n)

(n+1)p+1 - 1 - Σk=0...p-1 C(p+1,k)Sk(n) = C(p+1,p)Sp(n)

[ (n+1)p+1 - 1 - Σk=0...p-1 C(p+1,k)Sk(n) ] / (p+1) = Sp(n)

See the derivation of the square pyramidal number formula for an example of this technique in use.


2. The infinite sum of the reciprocals of S₃(n) is 4π²/3 - 12. This can be proven as follows:

Σn=1...∞ 1/S₃(n) = Σn=1...∞ 4/[n²(n+1)²]

= Σn=1...∞ [-8/n + 8/(n+1) + 4/n² + 4/(n+1)²]

= -8Σn=1...∞[1/n - 1/(n+1)] + 4Σn=1...∞ 1/n² + 4Σn=2...∞ 1/n²

= -8Σn=1...∞[1/n - 1/(n+1)] + 8Σn=1...∞ 1/n²      - 4

The first part of this expression is a telescoping sum which reduces to -8. To evaluate the second part of the expression, you must use the fact that
Σn=1...∞ 1/n² = π²/6. Putting everything together gives you

-8 + 8π²/6 - 4

= 4π²/3 - 12 ≈ 1.15947253478581149


3. The infinite sum of the reciprocals of S₄(n), S₅(n), and S₆(n) are

Σn=1...∞ 1/S₄(n) ≈ 1.07383121333922776

Σn=1...∞ 1/S₅(n) = 60 - 4π² + 8π3tan(π3/2) ≈ 1.03507429369914524

Σn=1...∞ 1/S₆(n) ≈ 1.01807600128


© Had2Know 2010

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