Certain step in the induction proof $\sum\limits_{i=0}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ unclear

Question

It's about proving the following: $$\sum\limits_{i=0}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$

I understand every step in the master solution, however, I have no idea how one can know by intuition to transform the formula as follows:

$$\frac{2n^3+9n^2+13n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} = \frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$$

I've cheated a little and started backwards from the end to understand how those terms are concluded but I'd like to know what information I'm missing to form these terms by intuition.

Am I supposed to always have a look at what the final formula should look like so I can "work towards it"?

Answer 1

One way to arrive at the result with a bit more intuition, is to assume that $$\tag1\sum_{i=1}^n i^2=f(n)$$ where $$f(n)=an^3+bn^2+cn+d$$ for suitable coefficients $a,b,c,d$. Then from $f(n)-f(n-1)=n^2$ one obtains the equation $$\begin{align} n^2 &= a\cdot(n^3-(n-1)^3)+b\cdot(n^2-(n-1)^2)+c\cdot(n-(n-1)) \\&=a\cdot(3n^2-3n+1)+b\cdot(2n-1)+c\\&= 3a\cdot n^2+(2b-3a)\cdot n+(a-b+c)\end{align}$$ which holds for all $n$ iff $3a=1$, $2b-3a=0$, and $a-b+c=0$, which leads to $$\sum_{i=1}^n i^2=\frac13n^3+\frac12n^2+\frac16n+d$$ where we must adjust $d$ to match the initial conditions. It turns out that $d=0$ is good because we need $f(0)=0$.

Strictly speaking, we have derived necessary conditions for $a,b,c,d$ to hold if $(1)$ should hold for all $n$. But the way we made our transforms, it is immediatyly clear that for this choice of $a,b,c,d$ equation $(1)$ follows for all $n$ by induction.

Answer 2

At first the formulas for sums of squares/cubes... don't look obvious. Unlike the formula for sum of consecutive numbers, which is fairly easy to obtain using Gauss' logic, you can't find any obvious logic in the formulas for sums of squares and they look quite unnatural. But all those expressions are related to the Faulhaber's formula

Now since you know how the general formula should look like, you need to prove it for any natural numbers and here comes the induction. Having done the base and inductive hypothesis step you need to do the final third step. And for this step there isn't any particular way to do it, as long as you get the same general formula at the end. Here's few ways to prove your problem:

$$\sum_{i=1}^n i^2 + (n+1)^2 = \frac{n(n+1)(2n+1)}{6} + (n+1)^2$$ $$\sum_{i=1}^{n+1} i^2 = \frac{n(n+1)(2n+1) + 6(n+1)^2}{6}$$ $$\sum_{i=1}^{n+1} i^2 = \frac{(n+1)(n(2n+1) + 6(n+1))}{6}$$ $$\sum_{i=1}^{n+1} i^2 = \frac{(n+1)(2n^2+7n+6)}{6}$$ $$\sum_{i=1}^{n+1} i^2 = \frac{(n+1)(2n+3)(n+2)}{6}$$ $$\sum_{i=1}^{n+1} i^2 = \frac{(n+1)(2(n+1)+1)((n+1)+1)}{6}$$

Hence the proof. Also you can work backwards.

Answer 3

I usually do it like this:

You want to show that

$$\sum\limits_{i=0}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$

The base case it's okay. Now let's assume the formula is valid for $N$, we want to demonstrate the following, that is

$$\sum\limits_{i=0}^{N+1} i^2 = \frac{(N+1)(N+2)(2(N+1)+1)}{6} \ \ (1) $$

that is to say

$$\sum\limits_{i=0}^{N} i^2 + (N + 1)^2 = \frac{(N+1)(N+2)(2(N+1)+1)}{6}\ \ (2) $$

$\Rightarrow$

$$\frac{N(N+1)(2N+1)}{6} + (N+1)^2 = \frac{(N+1)(N+2)(2(N+1)+1)}{6}\ \ (3) $$

Now you dont need any "creativity", just show that the last equality is right by simplification and you're good

EDIT:

Just to make it clear, it is not a pitfall, because equalities are equivalence relations; that means that you can go from 1 to 2 to 3, as well as from 3 to 2 to 1. (since we went from 1 to 2 to 3 through a sequence of equalities)

Basically:

if you show that (3) is equivalent to $0=0$, which is True, then also (3) is true. Then also (2) is true. Then also (1) is true.

EDIT 2:

If you meant that you feel like "cheating" because you already know what the final formula look like, then the whole induction process is "cheating"

Induction can only be used to prove if a statement is right, but you have to "guess it" beforehand, somehow.

There are other methods that "builds" the final formula but induction is not one of them.

Answer 4

How about this.

To show that $$L(n)=\sum_{i=1}^{n}a_i=R(n)$$ holds for all $n$, you use induction. Step 1: Show that $$L(1)-R(1)=0.$$ Step 2: Show that $$L(n)=R(n)\quad\Rightarrow\quad L(n+1)=R(n+1).$$ Since $L(n+1)=L(n)+a_{n+1}$, we have $$L(n)=R(n)$$ $$L(n)+a_{n+1}=R(n)+a_{n+1}$$ $$L(n+1)=R(n)+a_{n+1}$$ so we need to show that $$R(n)+a_{n+1}=R(n+1)\qquad(1)$$ The standard method (let us call it method A) to show $(1)$ is to try to construct the form of $R(n+1)$ from $R(n)+a_{n+1}$. An alternative method (method B) is simply to check if $$R(n)+a_{n+1}-R(n+1)=0.$$ It is usually much easier to do method B, as you don't need to come up with any tricks from algebra. Just expand and simplify. If you get $0$, you're home free. In fact we can do better than this. It is sufficient to show that $$k\cdot[R(n)+a_{n+1}-R(n+1)]=0$$ for some nonzero $k$. This means that you can get rid of troublesome fractions on your way by multiplying by them. If you do that, you should write $\propto$ instead of $=$.

Answer 5

Your question can also be interpreted in this way: I have obtained the expression $2n^3+9n^2+13n+6$. How do I get from there to $(n+1)(n+2)(2n+3)$?

Although, as you said yourself, in this case it seems not that relevant. Since you are doing proof by induction you know what you want to obtain. But I will ignore this information, since this thing might be useful anyway.

When simplifying polynomial expressions, rational root theorem might be sometimes useful. In this particular case it says that if $2x^3+9x^2+13x+6$ has some rational root $x=p/q$, then necessarily $p\mid6$ and $q\mid2$. So you have only check the possibilities $p\in\{\pm1,\pm2,\pm3,\pm6\}$ and $q\in\{1,2\}$. So the candidates for rational roots are $\pm1,\pm2,\pm3,\pm6,\pm1/2,\pm3/2$. You do not even have to check all of them. As soon you can find one root, you can divide the expression to obtain the quadratic equation which you can solve (for example using discriminant).

For example in this case, if you find out that $-1$ is a root, then you get
$2x^3+9x^2+13x+6=(x+1)(2x^2+7x+6)$.
If you then solve $2x^2+7x+6$ and find out that it has roots $-2$ and $-3/2$, you have the factorization $2x^2+7x+6=2(x+2)(x+3/2)$ and $$2x^3+9x^2+13x+6=2(x+1)(x+2)(x+3/2)=(x+1)(x+2)(2x+3).$$