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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.

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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.

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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:

$$c=d$$ $$a - (n + 1)(n + 2)(n+3) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$$ $$a = \frac{n(n + 1)(n + 2)(n + 3)}{4} + \frac{4(n + 1)(n + 2)(n+3)}{4}$$ $$a = \frac{\bigg((n + 1)(n + 2)(n+3)\bigg)\bigg(n + 4\bigg)}{4}$$ $$a = b$$

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For alternative approach $$r(r+1)(r+2)=r(r+1)(r+2)\dfrac{[(r+3)-(r-1)]}{4}$$ $$r(r+1)(r+2)=\frac14r(r+1)(r+2)(r+3)-\frac14(r-1)r(r+1)(r+2),\,\,\,\,\,\color{Green}{\text{Telescope}}$$

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