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?

Since you want $n+1$ to appear as a factor at the end, I would just leave that as a factor: $$ \frac{n(n+1)(2n+1)+6(n+1)^2}6 = \frac {n(2n+1) + 6(n+1)} 6 (n+1). $$ That becomes $$ \frac {2n^2 + n + 6n + 6} 6 (n+1) = \frac{2n^2 + 7n+6} 6 (n+1) = \frac{(n+2)(2n+3)} 6 (n+1) $$

Oh goodness, you needn't so much work. All you need to do is this:

$$\begin{align}\frac{n(n+1)(2n+1)}6+(n+1)^2&=\frac{n(n+1)(2n+1)}6+\frac{6(n+1)^2}6\\&\tag 1=c\bigg(n(2n+1)+6(n+1)\bigg)\\&\tag2=c\bigg(2n^2+7n+6\bigg)\\&\tag3=c\bigg((n+2)(2n+3)\bigg)\end{align}$$

$(1)$ factor out $(n+1)/6$ and call it $c$.

$(2)$ expand the remaining stuff

$(3)$ factor the remaining stuff.

Here is one solution. Since each side of identity is a polynomial of third degree (you can reduce it to degree 2 by canceling $(n+1)$), you can just verify that it holds for $4$ (or just $3$ after cancellation) distinct values of $n$ say $n=1,0,-1,-2$ then you are done. Note that I have chosen specific values of $n$ in such a manner to make calculations easy (some terms vanish for $n=0,-1,-2$).

This is useful if expressions are complicated. See the end of this answer https://math.stackexchange.com/a/1814894/72031 where I use this technique to prove an identity involving polynomials of degree $4$.

Update: this is identical to Hagen von Eitzen's comment which I saw later. I am marking it as community wiki.