Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings


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*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $

*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$

*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$
Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next? 

Observe that:
$$(k+2)^2+\cdots+(2k+2)^2=$$$$\left[(k+1)^2+(k+2)^2+\cdots+(2k)^2\right]+(2k+1)^2+(2k+2)^2-(k+1)^2=$$$$f(k)+(2k+1)^2+(2k+2)^2-(k+1)^2$$
where $f(k)$ denotes the RHS of 2).
If the equality indeed holds then the outcome must be $f(k+1)$ and that can be checked.
