1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ No, it's not correct: It's off by a term involving $n$, after all (and even worse, it's never an integer). How did you get to that point? $\endgroup$
    – user296602
    Mar 30, 2016 at 3:46
  • $\begingroup$ Could you show your working so we can be more specific with our answers? $\endgroup$
    – shardulc
    Mar 30, 2016 at 3:51
  • $\begingroup$ How do you justify the first step? It makes no sense, you are no longer summing over anything... And where did $n$ go? $\endgroup$
    – Servaes
    Mar 30, 2016 at 4:35
  • $\begingroup$ You then proceed correctly, but make a misstep at the second to last step: \begin{equation} 6\frac{n^2+n}{2}-\frac{6n}{2}=\frac{6n^2+6n}{2}-\frac{6n}{2}=\frac{6n^2+6n-6n}{2}=\frac{6n^2}{2}=3n^2. \end{equation} ... I seem to be having some problems with LaTex myself. I hope the point is clear. $\endgroup$
    – Servaes
    Mar 30, 2016 at 4:37
  • $\begingroup$ Sorry the i & 3 on top of the sums were supposed to be to the right of the sum and there is supposed to be an N instead on top of both the sums instead(if that is what confused you). I just messed up the LaTeX. $\endgroup$
    – M.G
    Mar 30, 2016 at 4:40

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ You missed the $1$ in base case $\endgroup$ Mar 30, 2016 at 4:31
  • $\begingroup$ Your welcome.... $\endgroup$ Mar 30, 2016 at 4:36
1
$\begingroup$

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...

$\dots=6(n+1)-3+3n^2$ by induction hypothesis

Continuing to simplify:

$\dots = 6n+6-3+3n^2 = 3n^2+6n+3 = 3(n+1)^2$ which is exactly what we wanted to show.

Thus, by the principle of mathematical induction, the result is true for all $n\geq 1$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .