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

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 (1^p + 2^p + 3^p + ... + n^p) is denoted by S_p(n).

For positive integer values of p and n, the value of S_p(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 a^p to b^p

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

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),..., S_p-1(n), then you can find S_p(n) by expanding the sum Σ_j=1...n(1+j)^p+1 into individual power sums.

Σ_j=1...n (1+j)^p+1 = Σ_k=0...p+1[Σ_j=1...n C(p+1,k)j^k] = Σ_k=0...p+1 C(p+1,k)S_k(n)

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

Σ_j=1...n (1+j)^p+1 - S_p+1(n) = Σ_k=0...p C(p+1,k)S_k(n)

(n+1)^p+1 - 1 = Σ_k=0...p C(p+1,k)S_k(n)

(n+1)^p+1 - 1 - Σ_k=0...p-1 C(p+1,k)S_k(n) = C(p+1,p)S_p(n)

[ (n+1)^p+1 - 1 - Σ_k=0...p-1 C(p+1,k)S_k(n) ] / (p+1) = S_p(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:

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

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

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

= -8Σ_n=1...∞[1/n - 1/(n+1)] + 8Σ_n=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
Σ_n=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

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

Σ_n=1...∞ 1/S₅(n) = 60 - 4π² + 8π√3tan(π√3/2) ≈ 1.03507429369914524

Σ_n=1...∞ 1/S₆(n) ≈ 1.01807600128

© Had2Know 2010