How do I implement the $n=k+1$ step for proof by induction: $2+5+8+...+(3n-1) = {1\over 2} n(3n+1)$

Question

Attempting to use mathematical induction, I have first used the base case $n=1$

$(3-1)=0.5(3+1)$
$2={2\over 4}$
$2=2$

I believe I have done this correctly. Next I assume true where $n=k$

$2+5+8+...+(3k-1)={1\over 2}k(3k+1)$

Now I am supposed to show true for $n=k+1$ but here is where I am not sure because I don't know exactly how I can properly add $k+1$ to the left and right hand sides of the $n=k$ equation. This is what I have tried:

$2+5+8+...+(3k-1)+(3(k+1)-1)={1\over 2}k+1(3(k+1)+1)$

Have I done this correctly? I can follow the simpler examples of mathematical induction but I am absolutely stumped on how to go about doing this one.

Answer 1

The last equation is not correct. It should read

$$2+5+8+...+(3k−1)+(3(k+1)−1)=\frac{1}{2}(k+1)(3(k+1)+1).$$

Answer 2

So, we assume for our induction hypothesis that $2+5+8+\dots + (3k-1)=\frac{1}{2}k(3k+1)$

You already proved the base case.

We want to prove that it follows that $2+5+8+\dots+(3k-1)+(3(k+1)-1)=\frac{1}{2}(k+1)(3(k+1)+1)$. To do this, we start with one side of the expression and through a series of equalities and applying our induction hypothesis show that by transitivity it equals the other side.

$\begin{array}{rl}2+5+8+\dots+(3k-1)+(3(k+1)-1)&=^{\color{blue}{\text{I.H.}}}\frac{1}{2}k(3k+1) + (3(k+1)-1)\\ &=\frac{1}{2}(3k^2+k)+(3k+2)\\ &=\frac{1}{2}(3k^2+k)+\frac{1}{2}(6k+4)\\ &=\frac{1}{2}(3k^2+7k+4)\\&=\frac{1}{2}(k+1)(3k+4)\\ &=\frac{1}{2}(k+1)(3(k+1)+1)~~\square\end{array}$

It helps to be clear where exactly you used your induction hypothesis. In the work above it occurred in the very first line where I marked it with an $\color{blue}{\text{I.H.}}$ which allowed us to get rid of the summation, leaving the only the final term of the summation and the result of the sum of all of the earlier terms.

From there, it was a matter of algebraic manipulation. My thought process was that I wanted to expand everything out, combine them, and then refactor the expressions. There was that messy $\frac{1}{2}$ hanging around however and I preferred to keep the majority of my arithmetic done with integers instead of fractions so I opted to change the right parenthetical phrase to have an extra factor of $\frac{1}{2}$ out front by the technique of "multiplying by one" which in this case meant simultaneously multiplying by $2$ and $\frac{1}{2}$. This allowed the parenthetical phrases to be combined. Once combined you can use whatever technique you are comfortable with to factor these (e.g. the quadratic formula).

Finally, just a little more algebraic manipulation to get it into the exact form we were expecting to complete the induction step.