# Question #390aa

May 9, 2016

${h}_{n} = \sqrt{n + 1}$

#### Explanation:

We will proceed to show that ${h}_{n} = \sqrt{n + 1}$ using induction.

First, for our base case, note that by the Pythagorean theorem, ${h}_{1}^{2} = {1}^{1} + {1}^{1} = 2$, meaning ${h}_{1} = \sqrt{1 + 1}$, satisfying our equation.

Next, suppose that for some integer $k \ge 1$ we have ${h}_{k} = \sqrt{k + 1}$. We must show that ${h}_{k + 1} = \sqrt{\left(k + 1\right) + 1}$.

Indeed, as ${h}_{k + 1}$ is the hypotenuse of the right triangle with legs of lengths $1$ and ${h}_{k}$, we have

${h}_{k + 1}^{2} = {1}^{2} + {h}_{k}^{2} = 1 + {\sqrt{k + 1}}^{2} = \left(k + 1\right) + 1$

Taking the square root of both sides, we obtain

${h}_{k + 1} = \sqrt{\left(k + 1\right) + 1}$

as desired. Thus, by induction, ${h}_{n} = \sqrt{n + 1}$ for all $n \ge 1$.