수학적 귀납법

For each $n \in \mathbb{N}$, let $P( n )$ be, $2 + 5 + 8 + \cdots + (3n - 1) = \dfrac{{n( {3n + 1} )}}{2}$. When we use $n = 1$, the summation on the left side of the equation is 2, and the right side is $\dfrac{1(3 \cdot 1 + 1)}{2} = 2$. Therefore, $P( 1 )$ is true. For the inductive step, let $k \in \mathbb{N}$ and assume that $P \left( k \right)$ is true. Then,

$$2 + 5 + 8 + \cdots + \left( {3k - 1} \right) = \frac{{k \left( {3k + 1} \right)}}{2}$$

. We now add $3 \left( k + 1 \right) - 1$ to both sides of this equation. This gives

$$\begin{aligned}2 + 5 + 8 + \cdots + (3k - 1) & \\ + 3 \left( {k + 1} \right)-1 & = \frac{{k \left( {3k + 1} \right)}}{2} + \left( 3 \left( k + 1 \right) - 1 \right) \\ & = \frac{{k \left( {3k + 1} \right)}}{2} + \left( {3k + 2} \right)\end{aligned}$$

If we now combine the terms on the right side of the equation into a single fraction, we obtain

$$\begin{aligned}2 + 5 + \cdots + \left( {3k - 1} \right) + 3 \left( {k + 1} \right)-1 & = \frac{k \left( 3k + 1 \right) + 6k+4}{2} \\ & = \frac{ 3k^2 + 7k + 4}{2} \\ & = \frac{ \left( k + 1 \right) \left( 3k + 4 \right)}{2} \\ & = \frac{ \left( k + 1 \right) \left( 3 \left( k + 1 \right) + 1 \right)}{2}\end{aligned}$$

This proves that if $P\left( k \right)$ is true, then $P\left( {k + 1} \right)$ is true.