**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 = 5^{2}. That 6 + 10 = 16 = 4^{2} and 15 + 21 = 36 = 6^{2} can be illustrated in similar fashion. The general relationship can be expressed as *T _{n}* +

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 $$1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2} = {{n(n + 1)(n + 2)} \over 6}.$$