Extending al-Karaji's Work on Sums of Odd Powers of Integers - Conclusion and Exercises for the Reader

Conclusion

In this article we have introduced ideas from an article on sums of odd powers from the very first Turkish mathematics journal, Mebahis-i İlmiyye, by the editor of that journal, Vidinli Tevfik Pasha. Vidinli’s article explains and extends al-Karaji’s work. The mathematician Isaac Newton said, "If I have seen further it is only by standing on the shoulders of giants." If Vidinli has seen a little further, it is due to al-Karaji's work.

We leave the reader with a question: Can we apply the same method or similar methods to find the sums of other series? Some of the exercises below pose more specific versions of this question.

Exercises for the Reader

1. Verify algebraically the identity \[ {{(1 + 2 + 3 + \cdots + (n-1))}\times 2n + n^2 = n^3}\] given on page 2.


2. Complete the calculation of the formula for the sum of the first \( n\) seventh powers begun on page 5.


3. Complete the calculation of the formula for the sum of the first \( n\) ninth powers \[ {1^9 + 2^9 + 3^9 + \cdots + n^9} = {\frac{1}{10}}n^{10} + {\frac{1}{2}}n^9 + {\frac{3}{4}}n^8 - {\frac{7}{10}}n^6 + {\frac{1}{2}}n^4 - {\frac{3}{20}}n^2 \] begun at the bottom of page 5.


4. Suppose the lengths of the rectangles in al-Karaji's square from Figure 3 (page 2) were not \(1,2,3,\dots,n,\) but rather \(1,1,1,\dots,1.\) Apply the gnomon technique to the resulting diagram to derive the formula for the sum of consecutive odd numbers, \[1+3+5+\cdots +(2n-1) = {n^2}.\] You may compute the area of each gnomon either as the area of two rectangles together with the area of the square joining them or as the difference of the areas of two squares.


5. Suppose the lengths of the rectangles in al-Karaji's square from Figure 3 (page 2) were not \(1,2,3,\dots,n,\) but rather \(1,4,9,\dots,{n^2}.\) Apply the gnomon technique to the resulting diagram to derive the formula for the sum of the first \( n\) fifth powers, \[ {1^5 + 2^5 + 3^5 + \cdots + n^5} = {\frac{1}{6}}n^{6} + {\frac{1}{2}}n^5 + {\frac{5}{12}}n^4 - {\frac{1}{12}}n^2 .\] You may compute the area of each gnomon either as the area of two rectangles together with the area of the square joining them or as the difference of the areas of two squares, and you will need to use the formula for the sum of the first \( n\) third powers \[ {1^3 + 2^3 + 3^3 + \cdots + n^3} = {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2 = {\frac{1}{4}}n^{4} + {\frac{1}{2}}n^3 + {\frac{1}{4}}n^2 \] along the way.


6. Use a cube of side length \( n,\) divided into \( n\) nested three-dimensional gnomons, to derive the formula for the sum of the first \( n\) squares, \[ {1^2 + 2^2 + 3^2 + \cdots + n^2} = \frac{n(n+1)(2n+1)}{6} = {\frac{1}{3}}n^{3} + {\frac{1}{2}}n^2 + {\frac{1}{6}}n .\]


7. We extend the method of Exercise 6, a three-dimensional gnomon method inspired by al-Karaji's two-dimensional gnomon method, to four- and five-dimensional gnomons. Use a four-dimensional hypercube of side length \( n,\) divided into \( n\) nested four-dimensional gnomons, to derive the formula for the sum of the first \( n\) cubes, \[ {1^3 + 2^3 + 3^3 + \cdots + n^3} = {\frac{1}{4}}n^{4} + {\frac{1}{2}}n^3 + {\frac{1}{4}}n^2 .\] Use a five-dimensional hypercube of side length \( n,\) divided into \( n\) nested five-dimensional gnomons, to derive the formula for the sum of the first \( n\) fourth powers, \[ {1^4 + 2^4 + 3^4 + \cdots + n^4} = {\frac{1}{5}}n^{5} + {\frac{1}{2}}n^4 + {\frac{1}{3}}n^3 - {\frac{1}{30}}n .\] Explain how to extend this method to obtain a formula for the sum of the first \( n\) \( k\)th powers, \( k\ge 5 .\) Blaise Pascal (1623-1662) used a similar method for obtaining these formulas (see Beery 2009).


8. Use a cube of side length \( 1 + 2 + 3 + \cdots + n,\) divided into \( n\) nested three-dimensional gnomons of thicknesses \(1,2,3,\dots,n,\) to give a geometric interpretation of Vidinli's derivation on page 4 of the formula for the sum of the first \( n\) fifth powers.