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# 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*n*th 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 ^{p}

^{p}

*a*to

^{p}*b*(inclusive) is the difference between S

^{p}_{p}(b) and S

_{p}(a) plus the

*a*(You need to add

^{p}*a*because the sum is inclusive of the endpoints

^{p}*a*and

^{p}*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

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