<s>[INST] How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort? 
I will keep it short and take only an extract (most important part) of
  the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$

What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : /
If anyone wants see, here is my long solution which I'm not happy with:
$$\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}$$
 [/INST] Note that
$$n(2n+1)+6(n+1)=2n^2+7n+6=(n+2)(2n+3)$$
Multiplying both sides by $\dfrac {n+1}6$
$$\frac {n(n+1)(2n+1)}6+(n+1)^2=\frac {(n+1)(n+2)(2n+3)}6$$
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