I currently have to do this following proof using induction (base case, inductive hypothesis required):
$$\sum_{i=1}^n(6i-3)=3n^2, \forall n>1$$
I'm not really sure how to approach this question and currently i've ended up with $$=\frac{6n^2 - 5n}{2}$$ for the summation and don't really know where to go from here (or if that's even correct). Any help about where to go from here or what I did wrong is greatly appreciated! Thank you.
On
Base case:
For $n=1$ the identity clearly holds: $$\sum_{i=1}^n(6i-3)=6\times1-3=3=3\times1^2=3n^2.$$
Inductive hypothesis:
Suppose the identity holds for some natural number $n$. Then for $n+1$ we have
\begin{eqnarray*} \sum_{i=1}^{n+1}(6i-3) &=&(6(n+1)-3)+\sum_{i=1}^n(6i-3)\\ &=&(6(n+1)-3)+3n^2=3n^2+6n+3\\ &=&3(n^2+2n+1)=3(n+1)^2, \end{eqnarray*} which show that the identity then also holds for $n+1$.
Induction:
By induction the identity holds for all natural numbers.
Is it true for $n=1$?
$\sum\limits_{i=1}^1(6i-3) = 6\cdot 1 - 3 = 3 = 3\cdot 1^2$ Yes, it is true for $n=1$.
Now, supposing that it is true that $\sum\limits_{i=1}^n(6i-3)=3n^2$ for some $n\geq 1$, we are curious whether or not it follows that it will necessarily also be true for $n+1$
I.e. we want to show that $\sum\limits_{i=1}^n(6i-3)=3n^2\Rightarrow \sum\limits_{i=1}^{n+1}(6i-3)=3(n+1)^2$. To do so, we start at one side and manipulate it using a series of equalities, eventually at the very end arriving at the right hand side, showing the left side equals the right side. (do NOT begin with the equality you wish to prove and simply reach a tautology, that is circular logic and invalid)
$\sum\limits_{i=1}^{n+1}(6i-3) = 6(n+1)-3 + \sum\limits_{i=1}^{n}(6i-3)$ by simply adding the final term of the summation individually.
Now, we recognize the smaller summation and we know something about it based on our induction hypothesis...
Continuing to simplify:
Thus, by the principle of mathematical induction, the result is true for all $n\geq 1$