Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
@SithsAndGiggles
1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2

Answer 1

Okay I don't see a readily available counter-example. According to WA, the formula is correct, but you want to prove it's correct.

An induction proof consists of the following:
(1) base case
(2) induction step

The base case establishes that the formula, in this case
\[\sum_{i=1}^n(3i-2)^2=\frac{n(6n^2-3n-1)}{2}\]
holds for the first (base) index. What this means is we plug in \(n=1\):
\[\sum_{i=1}^1(3i-2)^2=(3\cdot1-2)^2=1^1=1\\
\frac{1(6\cdot1^2-3\cdot1-1)}{2}=\frac{6-3-1}{2}=\frac{2}{2}=1\]
So the base case holds. So far so good?

Answer 2

yes! im folowing

Answer 3

so thats it right?

Answer 4

oh noo theres the induction step

Answer 5

The induction step involves this reasoning: If the formula holds for an arbitrary \(n\), and if we can prove it holds for the next \(n\), then the formula holds in general.

In other words, if we assume the formula is correct for \(n=k\), and we use that assumption to show it holds for \(n=k+1\), then the formula is true for all \(n\).

Answer 6

In context: Assume the formula holds for \(n=k\). This means
\[1^2+4^2+\cdots+(3k-2)^2=\frac{k(6k^2-3k-1)}{2}\]
What we want to show:
\[1^2+4^2+\cdots+(3k-2)^2+(3(k+1)-2)^2=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]
Notice that we just add the \(k+1\)-th term to the left side and substitute \(k+1\) for \(n\) on the right side. Do you follow?

Answer 7

yes

Answer 8

is the ending -3(k+1-2? it cut off

Answer 9

so that's it? it is true? @SithsAndGiggles

Answer 10

No, we haven't shown anything yet. I'm only making the connection between the induction proof strategy and the actual facts we have to work with.

Answer 11

ohhh now i understand, i got a little confused

Answer 12

As for what was cut off:
\[1^2+4^2+\cdots+(3k-2)^2+(3(k+1)-2)^2\]is equal to\[\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]

Answer 13

To be precise, we want to prove it's equal.

Answer 14

To do that, we start with the left hand side:
\[\color{red}{1^2+4^2+\cdots+(3k-2)^2}+(3(k+1)-2)^2\]
Recall the assumption for the first \(k\) terms:
\[1^2+4^2+\cdots+(3k-2)^2=\frac{k(6k^2-3k-1)}{2}\]
Substituting, we have
\[\color{red}{\frac{k(6k^2-3k-1)}{2}}+(3(k+1)-2)^2\]

Answer 15

We want to be able to combine the terms so we can establish
\[\color{red}{\frac{k(6k^2-3k-1)}{2}}+(3(k+1)-2)^2=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]

Answer 16

It's all algebra from here on out.
\[\begin{align*}\frac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2&=\frac{1}{2}\left(k(6k^2-3k-1)+2(3(k+1)-2)^2\right)\\
&=\frac{1}{2}\left(6k^3-3k^2-k+2(3k+1)^2\right)\\
&=\frac{1}{2}\left(6k^3-3k^2-k+2(9k^2+6k+1)\right)\\
&=\frac{1}{2}\left(6k^3-3k^2-k+18k^2+12k+2)\right)\\
&=\frac{1}{2}\left(6k^3+15k^2+11k+2\right)
\end{align*}\]
So to sum up, we have to prove this equality:
\[\frac{1}{2}\left(6k^3+15k^2+11k+2\right)=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]
Notice we already have the factor of \(\dfrac{1}{2}\) on both sides, so it comes down to showing the numerators are the same.
\[6k^3+15k^2+11k+2\overbrace{=}^{\large ?}(k+1)(6(k+1)^2-3(k+1)-1)\]

Answer 17

gotcha

Answer 18

Let's pull out a factor of \(k+1\) from the left side via long division: