Mathematical induction example

[cptstubing]

Homework Statement

A step in this process of proving S_n: 1+4+7+...+(3n-2) = n(3n-1)/2
confuses me. I hope someone can clarify this for me.
I do not require the work done, I need clarification on a step only. Thanks!

Homework Equations

After assuming n=k, we say S_k: 1+4+7+...+(3k-2) = k(3k-1)/2
When assuming n=k+1, we say S_k+1: 1+4+7+...+(3k-2) + (3k+1) = k(3k-1)/2 + (3k+1)
The book states the reason for adding (3k+1) on both sides of the equation is because 3(k+1)-2 = 3k+1

The Attempt at a Solution

Why is this the case? What is 3(k+1)-2 ? I know it equals 3k+1 because 3 multiplied by (k+1) is 3k+3, then -2 makes it 3k+1. But where did 3(k+1)-2 suddenly come from? It seems arbitrary and is without explanation in the textbook.

 

[MostlyHarmless]

When using induction you are trying to show that the k^th case implies the k+1^th case. So after adding 3k+1 on both sides you get S_k + 3k+1=S_k+3(k+1)-2, which if you are to assume the inductive hypothesis, holds with n=k+1.

 

[LCKurtz]

Homework Statement

A step in this process of proving S_n: 1+4+7+...+(3n-2) = n(3n-1)/2
confuses me. I hope someone can clarify this for me.
I do not require the work done, I need clarification on a step only. Thanks!

Homework Equations

After assuming n=k, we say S_k: 1+4+7+...+(3k-2) = k(3k-1)/2

I would have said "By the induction hypotheses, for n=k we assume that 1+4+7+...+(3k-2) = k(3k-1)/2."

When assuming n=k+1, we say S_k+1: 1+4+7+...+(3k-2) + (3k+1) = k(3k-1)/2 + (3k+1)
The book states the reason for adding (3k+1) on both sides of the equation is because 3(k+1)-2 = 3k+1

Here it is better to say what we are to prove, which is that $S_{k+1}$ is true, which is:$$

1+4+7+...+ (3k-2)+(3(k+1)-2) = \frac{(k+1)(3(k+1)-1)} 2$$
This is gotten by just writing $S_n$ when $n=k+1$. Now if you simplify the last term on the left side, you will see why adding $3k+1$ to both sides of $S_k$ will make the left sides equal and, hopefully, the right side equal to what you want.

 

[cptstubing]

I got it now.
That term 3(k+1)-2 that I thought was totally random came from plugging in (k+1) into the k in (3k-2)
&*@^&$@^(!