Learning how to prove by mathematical induction for the first time So, I have to prove this by mathematical induction and I have never done it!
$$\sum_{i=1}^n i^2 = n(n+1)(2n+1)/6$$
However, I have learnt better the theory behind this way to prove statements and studied a couple of easy examples yesterday. I am trying this now but it may be I make some mistakes - perhaps, at the beginning - because I don't get an identity in the final passage. Is this the goal, right? Namely, two have 2 equal lines in the equation (?)
This is my attempt:
Let's prove the base case before with 2 (I choose 2 instead than 1 because with 1 the formula looks more trivial, in my opinion. I guess the choice of the number is arbitrary). Thus:
n=2 , i = sum of natural numbers, according to the rule, up to n, in this case, up to 2. I get:
$$1^2 + 2^2 = 2(2+1) (2*2+1) / 6$$
$$5 = 5$$  (I jumped a step in the computation here to be shorter)
So, the formula is true for 2.
Now, I have to introduce a general number k such that k is a natural number and, for underlying the progression of N, I write: k+1.
However, at this point I have my first doubt: how to write i^2?. Should I leave it like this and replacing only n by k+1?. I am trying to find a way for writig also i in terms of k+1 but I can't find it. So, for now I leave it like this and I work only on the right side of the formula.
As I have understood so far, I have to replace n by k+1 simply. So, I write only the right part of the formula in this way:
$$(k+1)*[(k+1)+1]*[2(k+1)+1] / 6$$
I compute the terms:
$$(k+1)*(k+2)*(2k+3) / 6$$
I expand:
$$k^2+2k+k+2*(2k+3) / 6$$
I sum like terms:
$$(k^2+3k+2)*(2k+3) / 6$$
Expand again:
$$2k^3+3k^2+6k^2+9k+4k+6 / 6$$
Sum like terms again:
$$(2k^3+9k^2+13k+6) / 6$$
Already in this passage I don't like this 13, prime number which doesn't allow me to simplify nicely, and which let me think that I made a mistake. I could simplify in this way:
$$k^3/3 + 3k^2/2 + 13k/6 + 1$$  but I don't end up with any similarity with the left side of the formula.
Could you give me some help, please? ...I am already tired...
UPDATE:
I hope you don't put this post as a duplicate for the fact is going to be proved the same statement, as I am trying also to learn how to write a proof with my own style. So, answers to this question may help me to achieve the goal, at least for this question. However, I read and I will also similar questions already posted in order to learn more about proof writing.
UPDATE: Can I write the proof in the following way? After proved for the base case, I write:
Let's assume that k=n and there exists a general number k+1 which appartains at N (I would like to condensate what you refer as steps 2 and 3) so that $$\sum_{i=1}^n i^2 = \sum_{i=1}^{i=k+1} i^2 = k(k+1)(2k+1)/6 + (k+1)^2$$ - I don't know how to put that 1 upon sigma
and then I simply proceed with the algebraic passages (?)
 A: Induction is a 3 step process.  The first step will always be the base case.  So, assuming induction on the natural numbers or some subset of the natural numbers, there will always be a least element that must satisfies the expression requiring proof.  For your case, you need to show that 
$$\sum_{i=1}^1{i^2}=1^2=1$$ and $$\frac{1(1+1)(2\cdot 1+1)}{6}=\frac{1\cdot 2\cdot 3}{6}=\frac6{6}=1$$
Thus, for $n=1$, this formula for the sum of natural number squares holds.
Now for induction, we ASSUME that the case where $n=k$ is true.  Notice here that $k$ is assumed to be ANY natural number bigger than 1 (since we wanted the lowest such number to be true as our base case).  Under this assumption then,
$$\sum_{i=1}^ki^2=1^2+2^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$$
Finally, the last step is to show that this particular formula holds when we replace $k$ with $k+1$.  This method is sometimes called the domino proof because if you consider assuming true a statement for some $n$ and it is then true for the next number, it is then true for all natural numbers, since our choice of $n$ is arbitrary.  To show the proof, start with 
$$\sum_{i=1}^{k+1}=1^2+2^2+...+k^2+(k+1)^2$$
Now we already have a formula for the sum of the first $k$ so we can replace that section with the formula:  in other words
$$\sum_{i=1}^{k+1}=1^2+2^2+...+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$
Now we finish.  We need to manipulate our right hand side expression to display the fact that we replaced $k$ with $k+1$, so our goal is to show that 
$$\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{[(k+1)[(k+1)+1)][2(k+1)+1]}{6}$$
It is easier not to simply multiply out.  Leave the LHS in factored form
$$\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$$
$$=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}$$
$$=\frac{(k+1)(2k^2+7k+6)}{6}$$
$$=\frac{(k+1)(2k^2+4k+3k+6)}{6}$$
$$=\frac{(k+1)[2k(k+2)+3(k+2)]}{6}$$
$$=\frac{(k+1)(k+2)(2k+3)}{6}$$
$$=\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}$$
This completes the proof.
