Chapter 9, Problem 9PS

To determine

To prove: The nth pentagonal number $P_n$ is given by $P_n=\frac{n\left(3n-1\right)}{2}$ .

Answer to Problem 9PS

The proof is given below.

Explanation of Solution

Proof:

Put $n=1$ in the given formula for $P_n$ .

  $\begin{array}{c}{P}_1=\frac{1\left(3\times 1-1\right)}{2}\\ =\frac{3-1}{2}\\ =1\end{array}$

This is true as $P_1$ represents the first pentagonal number 1.

Let the formula be true for $n=k$ , that is, $P_k=\frac{k\left(3k-1\right)}{2}$ represents the kth pentagonal number.

To prove that the formula holds for $n=k+1$ , proceed as follows:

  $\begin{array}{c}{P}_{k+1}=\frac{\left(k+1\right)\left(3\left(k+1\right)-1\right)}{2}\\ =\frac{\left(k+1\right)\left(3k+2\right)}{2}\\ =\frac{3k^2+2k+3k+2}{2}\\ =\frac{3k\left(k+1\right)}{2}+2\left(k+1\right)\end{array}$

Observe that $k\left(k+1\right)$ will always be an even number, irrespective if k is even or odd. This implies that the expression obtained for $P_{k+1}$ is completely divisible by 2, implying that $P_{k+1}$ is an integer representing the ${\left(k+1\right)}^{th}$ number in the sequence of pentagonal numbers.

Hence, by the principal of mathematical induction, the formula $P_n=\frac{n\left(3n-1\right)}{2}$ represents the nth pentagonal number for all values of $n\ge 1$ .