Proving the formula 3x + 6x + 9x + ... + nx = (3/2)nx(n+1)x using the Principle of Mathematical Induction (PMI)

Introduction:
The goal is to prove the given formula using the Principle of Mathematical Induction. PMI is a powerful technique used to prove statements about natural numbers. It involves two steps: the base case and the induction step.

Base Case:
We start by proving the formula for the smallest possible value of n, which is 1.

When n = 1, the formula becomes:
3x = (3/2)(1)(1+1)x
3x = (3/2)(1)(2)x
3x = 3x

The base case is true, as both sides of the equation are equal when n = 1.

Induction Hypothesis:
Assume that the formula holds for an arbitrary positive integer k, i.e., assume that 3x + 6x + 9x + ... + kx = (3/2)k(k+1)x is true.

Induction Step:
We need to prove that the formula also holds for the next integer, k+1.

Starting with the left-hand side of the equation:
3x + 6x + 9x + ... + kx + (k+1)x

Using the induction hypothesis, we can substitute the formula for the sum up to k:
(3/2)k(k+1)x + (k+1)x

Factoring out (k+1)x:
[(3/2)k(k+1) + (k+1)]x

Simplifying the expression:
[(3k^2 + 3k + 2k + 2)/2]x
[(3k^2 + 5k + 2)/2]x

Factoring the numerator:
[(3k^2 + 3k + 2k + 2)/2]x
[(k(3k + 3) + 2(3k + 1))/2]x
[(k+1)(3k + 2)/2]x

The right-hand side of the equation becomes:
(3/2)(k+1)(k+2)x

Since the right-hand side matches the simplified expression of the left-hand side, we have proven that if the formula holds for k, it also holds for k+1.

Conclusion:
By proving the base case and the induction step, we have proved that 3x + 6x + 9x + ... + nx = (3/2)nx(n+1)x for all positive integers n using the Principle of Mathematical Induction.