How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this 


  $$\sqrt{1+155}=2\sqrt{39}$$
  $$\sqrt{1+1555}=2\sqrt{389}$$
  $$\sqrt{1+15555}=2\sqrt{3889}$$
  $$\sqrt{1+155555}=2\sqrt{38889}$$
The formal statement is
$$\sqrt{1 + 10^k + 5 \sum_{j=0}^{k-1} 10^j} = 2\sqrt{3\cdot 10^{k-1} + 8 \sum_{j=1}^{k-2} 10^j + 9} \quad\forall k\ge 2$$
The base case can be shown by verifying $\sqrt{156} = \sqrt{4\cdot 39} = 2\sqrt{39}$. Now try induction (first you should eliminate the square root). We get
$$\begin{align*}
1 + 10^{k+1} + 5 \sum_{j=0}^k 10^j &amp;= 10^{k+1} + 1 + 5\cdot 10^k + 5\sum_{j=0}^{k-1} 10^j \\
&amp;\stackrel{\text{i.h.}}= 14 \cdot 10^k +4\cdot\left(3\cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9\right) \\
&amp;= 140 \cdot 10^{k-1} +4\cdot\left(3\cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9\right) \\
&amp;= 4 \cdot \left( 38 \cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9 \right) \\
&amp;= 4 \cdot \left( 3 \cdot 10^k + 8\cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9 \right) \\
&amp;= 4 \cdot \left( 3 \cdot 10^k + 8\sum_{j=1}^{k-1} 10^j + 9 \right)
\end{align*}$$
as claimed.
Well, 

$$\sqrt{155+1}=\sqrt{156}=\sqrt{4\times39}=\sqrt{4}\times\sqrt{39}=2\sqrt{39}$$
$$\sqrt{1555+1}=\sqrt{1556}=\sqrt{4\times389}=\sqrt{4}\times\sqrt{389}=2\sqrt{389}$$

And on..