Chapter F, Problem 40E

Prove formula (e) of Theorem 3 using the following method published by Abu Bekr Mohammed ibn Alhusain Alkar-chi in about ad 1010. The figure shows a square ABCD in which sides AB and AD have been divided into segments of lengths 1, 2, 3, …, n. Thus the side of the square has length n(n + 1)/2 so the area is [n(n + 1)/2]². But the area is also the sum of the areas of the n “gnomons” G₁, G₂, ..., G_n shown in the figure. Show that the area of G_i is i³ and conclude that formula (e) is true.

To determine

To show: The area of $G_i=i^3$ and conclude that ${\displaystyle \sum _{i=1}^n{i}^3}={\left[\frac{n\left(n+1\right)}{2}\right]}^2$.

Explanation of Solution

The area of $G_i$ is given by,

$\begin{array}{c}A={\left({\displaystyle \sum _{k=1}^{i}k}\right)}^2-{\left({\displaystyle \sum _{k=1}^{i-1}k}\right)}^2\\ ={\left[\frac{i\left(i+1\right)}{2}\right]}^2-{\left[\frac{i\left(i-1\right)}{2}\right]}^2\\ =\frac{i^2}{4}\left[{\left(i+1\right)}^2-{\left(i-1\right)}^2\right]\\ =\frac{i^2}{4}\left[\left(i+1+2i\right)-\left(i-2i+1\right)\right]\end{array}$

Simplify further as,

$\begin{array}{c}A=\frac{i^2}{4}\left[i+1+2i-i+2i-1\right]\\ =\frac{i^2}{4}\left[4i\right]\\ =i^3\end{array}$

Hence, the area is proved.

Thus, from the given information, the area of ABCD is given by,

${\displaystyle \sum _{i=1}^n{i}^3}={\left[\frac{n\left(n+1\right)}{2}\right]}^2$.