# Mathematical induction example

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1. Mar 30, 2015

### cptstubing

1. The problem statement, all variables and given/known data
A step in this process of proving Sn: 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!

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

3. 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 text book.

Last edited: Mar 30, 2015
2. Mar 30, 2015

### MostlyHarmless

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

3. Mar 30, 2015

### LCKurtz

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

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.

Last edited: Mar 30, 2015
4. Mar 31, 2015

### 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)
&*@^&\$@^(!