Inductive demonstration I have this:
$$1^2+2^2+3^2+\dots+n^2=\frac{n(n+1)(2n+1)}6$$
So I was suggested of doing this:
$$\begin{align}
(1^2+2^2+3^2+\dots+k^2)+(k+1)^2 
&= \frac{k(k+1)(2k+1)}6+(k+1)^2 \\
&= \frac{k(k+1)(2k+1)+\color{blue}{6(k+1)^2}}6 \\
&= \frac{(k+1)[k(2k+1)+6(k+1)]}6 \\
&= \frac{(k+1)[2k^2+7k+6]}6 \\
&= \frac{(k+1)(k+2)(2k+3)}6 \\
\end{align}$$
For making the prof, we suppose that the statement is true for any number.  (n=k) Then we can see that it is also true for any number k+1, adding this in both terms.  Every thing is ok here but then, like we can see in the right term (marked in blue)  appears some coefficient 6 of (k+1)^2 and I don't know it's origin. 
Can any one help me here?
The prof after say:
"Thus the left-hand side of (2) is equal to the right-hand side of (2). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integer n".
And also in this statement it is not clear to me why is true for n.

\begin{align*}
\frac{k(k+1)(2k+1)}6+(k+1)^2&=\frac{k(k+1)(2k+1)}6+\frac{6(k+1)^2}{6}\\
&=\frac{k(k+1)(2k+1)+6(k+1)^2}6
\end{align*}
For the last line,
$$\frac{(k+1)(k+2)(2k+3)}6=\frac{(k+1)[(k+1)+1][2(k+1)+1]}6$$
