Understanding Mathematical Induction problems I have problem to understand this 3 formulas, I am new in this type of problems.
I have to solve this problems by induction.
\begin{align}
\sum_{j = 1}^{j = n} j^3 &= \left(\frac{n(n + 1)}{2}\right) ^ 2  &\text{where } n \geq 1 \\
\sum_{j = 1}^{j = n} j(j + 1) &= \frac{1}{3}n(n + 1)(n + 2)  & \text{where } n \geq 1 \\
\sum_{j = 1}^{j = n} j(j!) &= (n + 1)! - 1 
\end{align}

The general procedure for a proof by induction is to first show that the base case satisfies that proposition. Then you assume that your proposition holds for some $n$ and show that from their you can get $n+1$ to work. 
I'll work the first the one and let you try the others using the same procedure. 
We need to show that when $n=1$ the statement is true. clearly $1^3 = (\frac{1*2}{2})^2$. 
Now we assume that the statement is true for some $n$. Then we compute:
$$\sum_{j=1}^{n+1}j^3 = \sum_{j=1}^{n}j^3 + (n+1)^3 = \left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3 $$
$$ = \frac{n^2(n^2+2n+1)}{4} + (n^3+3n^2+3n+1)$$
$$ = \frac{n^4+2n^3+n^2}{4} + \frac{4(n^3+3n^2+3n+1)}{4}$$
$$ = \frac{n^4+6n^3+13n^2 + 12n+4}{4} = \left(\frac{(n+1)(n+2)}{2}\right)^2$$
Therefore the proposition holds for $n+1$ as well and so must hold for all $n$.
Conceptually think about it as a bunch of dominoes falling over one at a time. The base case guarantees that the first domino falls over, and the step of showing that you can get $n+1$ from the statement being true for $n$ is saying that any domino that falls will hit the next one.
