# Extending al-Karaji's Work on Sums of Odd Powers of Integers - The Sum of the Fifth Powers

Author(s):
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

Vidinli then asks, “What happens if we apply the same procedure to find the differences of third powers of $S_n$?” His answer consists of the following work: \begin{align} S_n^{\,3} - S_{n-1}^{\,\,3} &= {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^3 - {\bigg[ {\frac{(n-1)n}{2}}\bigg]}^3 \\ &= {\frac{n^3}{8}} \Big({(n+1)}^3 - {(n-1)}^3 \Big) \\ &= {\frac{n^3}{8}}(n^3 + 3n^2 + 3n +1 - n^3 + 3n^2 - 3n + 1) \\ &= {\frac{n^3}{8}}(6n^2 + 2) \\ &= {\frac{1}{4}}(3n^5 + n^3) \\ &= {\frac{3}{4}}n^5 + {\frac{1}{4}}n^3 , \\ S_{n-1}^{\,\,3} - S_{n-2}^{\,\,3} & = {\frac{3}{4}}{(n-1)}^5 + {\frac{1}{4}}{(n-1)}^3 ,\\ S_{n-2}^{\,\,3} - S_{n-3}^{\,\,3} & = {\frac{3}{4}}{(n-2)}^5 + {\frac{1}{4}}{(n-2)}^3 , \\ \\ \\ \dots\dots\dots & \dots\dots\dots\dots\dots ,\\ \\ S_2^{\,3} - S_{1}^{\,3} & = {\frac{3}{4}}2^5 + {\frac{1}{4}}2^3 , \\ S_1^{\,3} - S_{0}^{\,3} & = {\frac{3}{4}}1^5 + {\frac{1}{4}}1^3 .\end{align}

When we add these equations, we obtain: ${S_n^{\,3}} = {\frac{3}{4}}\Big(1^5 + 2^5 + 3^5 + \cdots + n^5\Big) + {\frac{1}{4}}\Big(1^3 + 2^3 + 3^3 + \cdots + n^3\Big) ,$ or ${{(1+2+3+\cdots +n)}^3} = {\frac{3}{4}}\Big({1^5 + 2^5 + 3^5 + \cdots + n^5}\Big)$ $+ {\frac{1}{4}}\Big({1^3 + 2^3 + 3^3 + \cdots + n^3}\Big) ,$ as can be seen in Figure 7.

Figure 7. An equation involving the sum of the fifth powers (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

We already know that $1^3 + 2^3 + 3^3 + \cdots + n^3 = {{(1+2+3+\cdots +n)}^2} = {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2$ and ${1 + 2 + 3 + \cdots + n} = {\frac{n(n+1)}{2}}.$ When we substitute these identities into the equation in (or just above) Figure 7, we obtain: ${\bigg[ {\frac{n(n+1)}{2}}\bigg]}^3 = {\frac{3}{4}}\Big(1^5 + 2^5 + 3^5 + \cdots + n^5\Big) + {\frac{1}{4}}{\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2 .$

If we then re-arrange this equation, we can find a formula for the sum of the fifth powers: ${1^5 + 2^5 + 3^5 + \cdots + n^5} = {\frac{4}{3}}\Bigg({{\bigg[ {\frac{n(n+1)}{2}}\bigg]}^3 - {\frac{1}{4}}{\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2}\Bigg)$ or ${1^5 + 2^5 + 3^5 + \cdots + n^5} = {\frac{1}{6}}n^6 + {\frac{1}{2}}n^5 + {\frac{5}{12}}n^4 - {\frac{1}{12}}n^2 .$

Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University), "Extending al-Karaji's Work on Sums of Odd Powers of Integers - The Sum of the Fifth Powers," Convergence (August 2011), DOI:10.4169/loci003725