Sums of Powers of Positive Integers - Solutions to Exercises 1-3

Exercise 1.

Figure 12. The sum of the 4^th and 5^th triangular numbers is the 5^th square number.

Figure 12 illustrates that the sum of the 4^th and 5^th triangular numbers is the 5^th square number (or the square of side length 5), or that 10 + 15 = 25 = 5². That 6 + 10 = 16 = 4² and 15 + 21 = 36 = 6² can be illustrated in similar fashion. The general relationship can be expressed as T_n + T_n+1 = (n+1)², where T_n is the nth triangular number, or, since T_n = n(n + 1)/2, as $${{n(n + 1)} \over 2} + {{(n + 1)(n + 2)} \over 2} = (n + 1)^2.$$

The latter equation can be checked easily using algebra.

Exercise 2.

Figure 13. The first four triangular numbers represented using squares

Figure 14. The identity \( 1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2}\) for n = 4 (left),
and the identity from Exercise 1 for n = 3 (right)

Exercise 3. Six pyramids constructed of 1 + 3 + 6 + 10 = 20 cubes each can be fitted together to form a rectangular solid of dimensions 4 x 5 x 6, or 120 cubes in all, as shown in Figure 15. This construction illustrates that 6(1 + 3 + 6 + 10) = 4 x 5 x 6 or that \( 1 + 3 + 6 + 10 = \frac{4\times 5\times 6}{6}.\) In general, we have $$6\left( {1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2}} \right) = n(n + 1)(n + 2)$$ or $$1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2} = {{n(n + 1)(n + 2)} \over 6}.$$