TITLE: How can I prove $\sqrt{(111...)+(55...)^2}=5...6$
QUESTION [8 upvotes]: The formula in my question can be illustrated as follow:

$$\sqrt{11+5^2}=6$$
  $$\sqrt{111+55^2}=56$$ $$\sqrt{1111+555^2}=556$$ $$\sqrt{11111+5555^2}=5556$$ and so on

How can I prove the general formula 

$$\sqrt{\underbrace{11\ldots 1}_{n+1\text{ times}} + (\underbrace{55\ldots 5}_{n\text{ times}})^{2}} = (\underbrace{55\ldots 5}_{n-1\text{ times}}6)^{2}$$

REPLY [12 votes]: These equalities are just instances of
$$\sqrt{(2n+1)+n^2}=n+1.$$