# Sums of Powers of Positive Integers - Solutions to Exercises 4-6

Author(s):
Janet Beery (University of Redlands)

Exercise 4. For n = 3, three shells constructed of 6 12 = 6 cubes, 6 22 = 24 cubes, and 6 32 = 54 cubes fit together to form a 3 x 4 x 7 rectangular solid, as shown in Figure 16A. This construction illustrates that $6\left(1^2 + 2^2 + 3^2\right) = {3\cdot 4\cdot 7},$ or $1^2 + 2^2 + 3^2 = {{3\cdot 4\cdot 7} \over 6},$ or, more generally, $1^2 + 2^2 + 3^2 + \cdots + n^2 = {{n(n + 1)(2n + 1)} \over 6}$ for any positive integer n.

Nilakantha reasoned that the outside shell contained 6 32 = 54 cubes as follows (see Figure 16B):

Exercise 5. For n = 3, three shells constructed of 6 (1 2)/2 = 6 cubes, 6 (2 3)/2 = 18 cubes, and 6 (3 4)/2 = 36 cubes fit together to form a 3 x 4 x 5 rectangular solid, as shown in Figure 17A. This construction illustrates that $6\left(1 + 3 + 6\right) = {3\cdot 4\cdot 5},$ or $1 + 3 + 6 = \frac {3\cdot 4\cdot 5}{6},$ or, more generally, $1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2} = {{n(n + 1)(n + 2)} \over 6}$ for any integer n.

Nilakantha may have reasoned that the outside shell contained 6 (3 4)/2 = 36 cubes as follows (see Figure 17B):

Exercise 6. They may have made the following computations.

1 + 3 = 4 = 22

3 + 6 = 9 = 32

6 + 10 = 16 = 42

10 + 15 = 25 = 52

15 + 21 = 36 = 62

The general relationship can be expressed as

Tn + Tn+1 = (n+1)2,

where Tn is the nth triangular number, or as

${{n(n + 1)} \over 2} + {{(n + 1)(n + 2)} \over 2} = (n + 1)^2.$

The latter equation can be checked easily using algebra.

Janet Beery (University of Redlands), "Sums of Powers of Positive Integers - Solutions to Exercises 4-6," Convergence (July 2010), DOI:10.4169/loci003284