# Can someone check this proof for me, as I have looked over it, and I am not sure if it correct?

## $1 + 4 + 7. . . \left(3 n - 2\right) = \frac{n \left(3 n - 1\right)}{2}$ $\left(3 n - 2\right) = \frac{n \left(3 n - 1\right)}{2}$ Basis case: $n = 1$ $\left(3 \times 1\right) - 2 = \frac{1 \times \left(3 \times 1\right)}{2}$ $1 = 1$ Assume that $n = k$: $\left(3 k - 2\right) = \frac{k \left(3 k - 1\right)}{2}$ Prove $n = k + 1$ $\left(3 \left(k + 1\right)\right) - 2 = \frac{\left(k + 1\right) 3 \left(k + 1\right) - 1}{2}$ $3 k - 1 = \frac{\left(k + 1\right) \left(3 k\right)}{2}$ $3 k - 1 = \frac{\left(3 {k}^{2} + 3 k\right)}{2}$ $3 k - 1 = \frac{k \left(3 k + 3\right)}{2}$ I have looked up the answer, and it looks different, but the result looks similar to the result of the original problem.

Oct 25, 2016

I think your first eqaulity is wrong. Therefore the second derivation is wrong.

#### Explanation:

Assume n=5 in the fist equation. The left side is the become 13. But right side is 5x7=35. This indicate that the beginning equation is wrong.

Oct 25, 2016

See Below

#### Explanation:

Here's roughly how it should have looked

Claim: ${\sum}_{i = 1}^{n} \left(3 i - 2\right) = 1 + 4 + 7 + \ldots + \left(3 n - 2\right) = \frac{n \left(3 n - 1\right)}{2}$

Proof: (By Induction)

Base Case: For $n = 1$, we have

${\sum}_{i = 1}^{1} \left(3 i - 2\right) = \left(3 \left(1\right) - 2\right)$

$= 1$

$= \frac{2}{2}$

$= \frac{1 \left(3 \left(1\right) - 1\right)}{2}$

Notice that we start at one side of the equality and arrive at the other, rather than starting from both at the same time and arriving at a tautology. This is typically preferable, and prevents issues such as $- 1 = 1 \implies {\left(- 1\right)}^{2} = {1}^{2} \implies 1 = 1$

Induction Hypothesis: Suppose, for some integer $k \ge 1$, we have that

${\sum}_{i = 1}^{k} \left(3 i - 2\right) = 1 + 4 + 7 + \ldots + \left(3 k - 2\right) = \frac{k \left(3 k - 1\right)}{2}$

Notice that we do not assume that $k = n$. Rather, we assume that the claim is true for some arbitrary $k$ which is greater than or equal to our base case. This is what allows us to say "the base case is true, so the next case is true, so the next case is true, so... so all cases are true." Also, be careful to copy the claim correctly. Our claim is regarding a sum of integers up to $3 k - 2$, not $3 k - 2$ itself.

Inductive Step: We wish to show that the claim holds true for $k + 1$, that is, that ${\sum}_{i = 1}^{k} \left(3 i - 2\right) = \frac{\left(k + 1\right) \left(3 \left(k + 1\right) - 1\right)}{2}$
Indeed:

${\sum}_{i = 1}^{k + 1} \left(3 i - 2\right) = 1 + 3 + \ldots + \left(3 \left(k + 1\right) - 2\right)$

$= {\sum}_{i = 1}^{k} \left(3 i - 2\right) + \left(3 k + 1\right)$

$= \frac{k \left(3 k - 1\right)}{2} + \left(3 k + 1\right)$
(Here is where we used our induction hypothesis)

$= \frac{3 {k}^{2} - k}{2} + \frac{6 k + 2}{2}$

$= \frac{3 {k}^{2} + 5 k + 2}{2}$

$= \frac{\left(k + 1\right) \left(3 k + 2\right)}{2}$

$= \frac{\left(k + 1\right) \left(3 \left(k + 1\right) - 1\right)}{2}$

As we have supposed true for $k$ and shown true for $k + 1$, the result follows by induction. Thus, for all $n \ge 1$, we have

${\sum}_{i = 1}^{n} \left(3 i - 2\right) = \frac{n \left(3 n - 1\right)}{2} \text{ }$

Notice again that we carefully matched the format of the original claim, and started from one side of the equality and showed the other, rather than starting from both sides and working our way to a tautology. Doing so is better left as scratch work.

Once you are more comfortable with proofs by induction, as well as proofs in general, there are some liberties you can take in the form and what you add/say. As someone just learning, though, it is better to stick to the format closely.

Jan 6, 2017

$\textcolor{red}{\text{Checking that the summation equation is correct.}}$

Given:" "1+4+7+...+(3n-2))

Let the sum be $s$

Let any place count be $i$

Let the ${i}^{\text{th}}$ term be ${a}_{i}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Then we have:

$s = {\sum}_{i = 1 \to n} {a}_{i} = {\sum}_{i = 1 \to n} \left(3 i - 2\right)$

$s = 3 {\sum}_{i = 1 \to n} i - {\sum}_{i = 1 \to n} 2$

but ${\sum}_{i = 1 \to n} 2 = 2 n$ giving

$s = 3 {\sum}_{i = 1 \to n} i - 2 n$

But $3 {\sum}_{i = 1 \to n} i = 3 \left(n \times \text{mean value}\right)$ giving:

$s = 3 \times \frac{n \left(1 + n\right)}{2} - 2 n$

$s = \frac{3 n}{2} \left(1 + n\right) - 2 n$

$s = \frac{3 n + 3 {n}^{2}}{2} - \frac{4 n}{2}$

$s = \frac{3 {n}^{2} + 3 n - 4 n}{2}$

s= (3n^2-n)/2" "=" "(n(3n-1))/2 color(red)(larr" matches the question")
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Check}}$

$1 + 4 + 7 + 10 = 22$

$\frac{4}{2} \left(3 \times 4 - 1\right) = 2 \times 11 = 22 \textcolor{red}{\leftarrow \text{ works for these numbers}}$