Mathematical Induction Proof Question dealing with integers How would you use mathematical induction to prove that 
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n \cdot (n + 1) \cdot (n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$
I tried proving the base case of $n = 1$ but the left half is much larger than the right half after computing $n = 1$. Can someone please solve this problem and explain how and why the solution is correct? I've been stuck on it for hours.
 A: The base case for $n=1$ is actually quite easy, since:
$\frac{1\cdot 2\cdot 3 \cdot 4}{4} = 1\cdot 2 \cdot 3$
Assuming that it holds for $n$, we check $n+1$:
$1\cdot 2\cdot 3 + \cdots + n(n+1)(n+2) + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)}{4} + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3)}{4} = \frac{(n+1)(n+2)(n+3)(n+4)}{4}$
which is what we set out to prove.
A: You want to prove
$$\underbrace{1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + (n + 1)(n + 2)(n+3)}_a = \underbrace{\frac{(n + 1)(n + 2)(n + 3)(n+4)}{4}}_b$$
and you are allowed to assume
$$\underbrace{1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + n(n + 1)(n + 2)}_c = \underbrace{\frac{n(n + 1)(n + 2)(n + 3)}{4}}_d$$
So you want to prove $a=b$ given that $c=d$.  First notice the relationship between $c$ and $a$:
$$c+(n + 1)(n + 2)(n+3) = a$$
Now that you have some information, you can prove $a=b$ starting with our assumption:
$$c=d$$
$$a - (n + 1)(n + 2)(n+3) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$$
$$a  = \frac{n(n + 1)(n + 2)(n + 3)}{4} + \frac{4(n + 1)(n + 2)(n+3)}{4}$$
$$a  = \frac{\bigg((n + 1)(n + 2)(n+3)\bigg)\bigg(n + 4\bigg)}{4}$$
$$a  = b$$
A: For alternative approach $$r(r+1)(r+2)=r(r+1)(r+2)\dfrac{[(r+3)-(r-1)]}{4}$$
$$r(r+1)(r+2)=\frac14r(r+1)(r+2)(r+3)-\frac14(r-1)r(r+1)(r+2),\,\,\,\,\,\color{Green}{\text{Telescope}}$$
