The sum of other powers

October 21, 2012

Further to the derivation of the sum of squares, here are the derivations for the sum of cubes and the sum of the fourth powers. The same technique may be used for deriving the sum of higher powers.

Sum of cubes

Let \(S_3\) be the sum of cubes of the first \(n\) natural numbers, such that \[S_3 = 1^3 + 2^3 + 3^3 + \cdots + n^3.\]

Our aim is to derive a closed form formula for \(S_3\) in terms of \(n\). We have \[\begin{aligned} (x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1. \end{aligned}\]

Using this, we write the following sequence of equations where \(1 \le x \le n\): \[\begin{aligned} (1+1)^4 & = 1^4 + 4.1^3 + 6.1^2 + 4.1 + 1\\ (2+1)^4 & = 2^4 + 4.2^3 + 6.2^2 + 4.2 + 1\\ (3+1)^4 & = 3^4 + 4.3^3 + 6.3^2 + 4.3 + 1\\ & \cdots\\ (n+1)^4 & = n^4 + 4n^3 + 6n^2 + 4n + 1 \end{aligned}\]

If we add up each of the six columns separately, we get \[\begin{aligned} (1+1)^4 + (2+1)^4 + \cdots + (n+1)^4 = \, & 1^4 + 2^4 + 3^4 + \cdots + n^4\\ & + 4(1^3 + 2^3 + 3^3 + \cdots + n^3)\\ & + 6(1^2 + 2^2 + 3^2 + \cdots + n^2)\\ & + 4(1 + 2 + 3+ \cdots +n)\\ & + 1 + 1 + 1 + \cdots + 1. \end{aligned}\]

After subtracting the fourth powers on both sides of the equation, and substituting the closed form for the sum of the lower powers, we get \[\begin{aligned} (n+1)^4 = 1 + 4S_3 + 6\left[\frac{n(n+1)(2n+1)}{6}\right] + 4\left[\frac{n(n+1)}{2}\right] + n \end{aligned}\]

or, \[\begin{aligned} n^4 + 4n^3 + 6n^2 + 4n + 1 = 1 + 4S_3 + (2n^3 + 3n^2 + n) + (2n^2 + 2n) + n,\\ \end{aligned}\]

which, after further simplification gives

\[\begin{aligned} S_3 = \frac{n^4+2n^3+n^2}{4} = \left[\frac{n(n+1)}{2}\right]^2 \end{aligned}\]

Sum of the fourth powers

Let \(S_4\) be the sum of the fourth powers of the first \(n\) natural numbers, such that \[S_4 = 1^4 + 2^4 + 3^4 + \cdots + n^4.\]

Our aim is to derive a closed form formula for \(S_4\) in terms of \(n\). We have \[\begin{aligned} (x+1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1. \end{aligned}\]

Using this, we write the following sequence of equations where \(1 \le x \le n\): \[\begin{aligned} (1+1)^5 & = 1^5 + 5.1^4 + 10.1^3 + 10.1^2 + 5.1 + 1\\ (2+1)^5 & = 2^5 + 5.2^4 + 10.2^3 + 10.2^2 + 5.2 + 1\\ (3+1)^5 & = 3^5 + 5.3^4 + 10.3^3 + 10.3^2 + 5.3 + 1\\ & \cdots\\ (n+1)^5 & = n^5 + 5.n^4 + 10n^3 + 10n^2 + 5n + 1 \end{aligned}\]

If we add up each of the seven columns separately, we get \[\begin{aligned} (1+1)^5 + (2+1)^5 + \cdots + (n+1)^5 = \, & 1^5 + 2^5 + 3^5 + \cdots + n^5\\ & + 5(1^4 + 2^4 + 3^4 + \cdots + n^4)\\ & + 10(1^3 + 2^3 + 3^3 + \cdots + n^3)\\ & + 10(1^2 + 2^2 + 3^2 + \cdots + n^2)\\ & + 5(1 + 2 + 3+ \cdots +n)\\ & + 1 + 1 + 1 + \cdots + 1. \end{aligned}\]

After subtracting the fifth powers on both sides of the equation, and substituting the closed form for the sum of the lower powers, we get \[\begin{aligned} (n+1)^5 = 1 + 5S_4 + 10\left[\frac{n(n+1)}{2}\right]^2 + 10\left[\frac{n(n+1)(2n+1)}{6}\right] + 5\left[\frac{n(n+1)}{2}\right] + n \end{aligned}\]

or, \[\begin{aligned} n^5 + 5n^4 + 10n^3 + 10n^2 + 5n + 1 = \, & 1 + 5S_4 + \frac{5}{2}(n^4+2n^3+n^2)\\ & + \frac{5}{3}(2n^3 + 3n^2 + n) + \frac{5}{2}(n^2 + n) + n,\\ \end{aligned}\]

which, after further simplification gives \[\begin{aligned} S_4 = \frac{6n^5 + 15n^4 + 10n^3 - n}{30} = \frac{n}{30}(6n^4 + 15n^3 + 10n^2 - 1) \end{aligned}\]

Sum of the \(n\)th powers

Using binomial coefficients, we generalise our technique as follows: \[(n+1)^k = \binom{k}{0} + \binom{k}{1}S_{k-1} + \binom{k}{2}S_{k-2} + \cdots + \binom{k}{k}S_0\textrm{, where }S_j = \sum_{x=1}^nx^j.\]

Using this equation, we calculate the value of \(S_{k-1}\) recursively by applying the sum of the lower powers as follows: \[\begin{aligned} S_{k-1} = \frac{(n+1)^k - 1 - \displaystyle\sum\limits_{i=0}^{k-2} \binom{k}{k-i}S_i}{k}\textrm{, and }S_0 = n. \end{aligned}\]


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