induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$

Question

I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.

$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$

1st attempt: \begin{align*} & = (6n - 3) + 3n^2\\ & = 3n^2 + 6n - 3\\ & = (3n^2 + 5n - 4) + (n + 1) \end{align*}

That seems way wrong lol ^^^

2nd attempt: \begin{align*} f(n) & = 3 + 9 + 15 + ... + (6n - 3)\\ f(n + 1) & = 6(n + 1) - 3\\ f(n + 1) & = 6(n - 3) + 6(n + 1) - 3\\ & = ? \end{align*}

I don't know I feel like I'm headed in the wrong direction

I guess another attempt I have would be: \begin{align*} f(n) & = 3n^2\\ f(n+1) & = 3(n + 1)^2\\ & = 3(n^2 + 2n + 1)\\ & = 3n^2 + 6n + 3\\ & = f(n) + (6n + 3) \end{align*}

Answer 1

An other way

$$\sum_{k=1}^n(6k-3)=6\sum_{k=1}^n k-3n=6\cdot\frac{n(n+1)}{2}-3n=3n^2.$$

The only proof you need to do (by induction if you need to use induction) is that $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$

Answer 2

Without induction you can do it easily taking three common so $3(1+3+..(2n-1))$ now we know that sum of odd terms is a perfect square so we can write as $3n^2$ or sum of AP=$\frac{n}{2}(2+(n-1)2)=n^2$

Answer 3

If you insist on induction :

Base case $n=1\ :\ 6\times 1-3=3=3\times 1^2$

If you assume $$3+9+15+...+(6n-3)=3n^2$$

You can conclude

$$3+9+15+...+(6n-3)+(6n+3)=3n^2+6n+3=3(n+1)^2$$

completing the proof.

Answer 4

Using summation notation gives $$3 + 9 + 15 + \cdots + 6n - 3 = \sum_{k = 1}^n (6k - 3)$$

Let $P(n)$ be the statement $$\sum_{k = 1}^{n} (6k - 3) = 3n^2$$

To prove this statement by induction, we must show that $P(1)$ holds and that whenever $P(m)$ holds for some positive integer $m$, then $P(m + 1)$ holds since we then have the chain of implications $$P(1) \implies P(2) \implies P(3) \implies \cdots$$

Proof. Let $n = 1$. Then $$\sum_{k = 1}^{1} (6k - 3) = 6 \cdot 1 - 3 = 6 - 3 = 3 = 3 \cdot 1 = 3 \cdot 1^2$$ so $P(1)$ holds.

Since $P(1)$ holds, we may assume there exists a positive integer $m$ such that $P(m)$ holds. Then $$\color{green}{\sum_{k = 1}^{m} (6k - 3) = 3m^2}$$

Let $n = m + 1$. Then \begin{align*} \sum_{k = 1}^{m + 1} (6k - 3) & = \color{green}{\sum_{k = 1}^{m} (6k - 3)} + [6(m + 1) - 3] && \text{by definition}\\ & = \color{green}{3m^2} + 6(m + 1) - 3 && \text{by the induction hypothesis}\\ & = 3m^2 + 6m + 6 - 3\\ & = 3m^2 + 6m + 3\\ & = 3(m^2 + 2m + 1)\\ & = 3(m + 1)^2 \end{align*} Hence, $P(m) \implies P(m + 1)$.

Since $P(1)$ holds and $P(m) \implies P(m + 1)$ for each positive integer $m$, $P(n)$ holds for all positive integers.$\blacksquare$