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

Question

How can I prove this

$$\sqrt{1+155}=2\sqrt{39}$$ $$\sqrt{1+1555}=2\sqrt{389}$$ $$\sqrt{1+15555}=2\sqrt{3889}$$ $$\sqrt{1+155555}=2\sqrt{38889}$$

Answer 1

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 $$\begin{align*} 1 + 10^{k+1} + 5 \sum_{j=0}^k 10^j &= 10^{k+1} + 1 + 5\cdot 10^k + 5\sum_{j=0}^{k-1} 10^j \\ &\stackrel{\text{i.h.}}= 14 \cdot 10^k +4\cdot\left(3\cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9\right) \\ &= 140 \cdot 10^{k-1} +4\cdot\left(3\cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9\right) \\ &= 4 \cdot \left( 38 \cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9 \right) \\ &= 4 \cdot \left( 3 \cdot 10^k + 8\cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9 \right) \\ &= 4 \cdot \left( 3 \cdot 10^k + 8\sum_{j=1}^{k-1} 10^j + 9 \right) \end{align*}$$ as claimed.

Answer 2

Hint:
Using a loose notation, $$1+155\ldots55=155\ldots56=2\cdot77\ldots78=4\cdot38\ldots89.$$ Now, take the square root of both sides and simplify.

Answer 3

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..

Answer 4

Square both sides, and you get $$4\cdot 3\overset{n\ times}{8\cdots8}9 = 4(9+8\sum_{k=1}^n 10^k+3\cdot 10^{n+1})= 36+32\sum_{k=1}^n 10^k +12\cdot 10^{n+1}= $$ $$=6+3\cdot 10+ 2\sum_{k=1}^n 10^k+3\sum_{k=1}^n 10^{k+1}+2\cdot 10^{n+1}+10^{n+2}= $$

$$= 6+ 2\sum_{k=1}^{n+1} 10^k+ 3\sum_{k=1}^{n+1} 10^k +10^{n+2}= 6+5\sum_{k=1}^{n+1} 10^k+10^{n+2}=1\overset{n+1\ times}{5\cdots 5}6= 1\overset{n+2\ times}{5\cdots 5}+1$$

Answer 5

You're essentially asking to prove $\sqrt{1+10^{n+1}+5\cdot\sum\limits_{k=0}^{n}10^k}=2\sqrt{1+3\cdot10^n+8\cdot\sum\limits_{k=0}^{n-1}10^k}$

So it's sufficient to prove:

• $1+10^{n+1}+5\cdot\sum\limits_{k=0}^{n}10^k=4\cdot(1+3\cdot10^n+8\cdot\sum\limits_{k=0}^{n-1}10^k)$
• $1+10^{n+1}+5\cdot\sum\limits_{k=0}^{n}10^k=4+12\cdot10^n+32\cdot\sum\limits_{k=0}^{n-1}10^k$
• $5\cdot\sum\limits_{k=0}^{n}10^k=3+2\cdot10^n+32\cdot\sum\limits_{k=0}^{n-1}10^k$
• $5\cdot10^n=3+2\cdot10^n+27\cdot\sum\limits_{k=0}^{n-1}10^k$
• $3\cdot10^n=3+27\cdot\sum\limits_{k=0}^{n-1}10^k$
• $10^n=1+9\cdot\sum\limits_{k=0}^{n-1}10^k$
• $10^n=10^n$

Answer 6

The expressions on the right are $$ \sqrt{4\left( 3\cdot 10^{k-1} + \frac{8}{9}\cdot 10^{k-1} - \frac{80}{9} + 9\right) }$$ for $k \geq 2$. I'm going to compare that to the expressions on the left, which have $k$ 5's. $$ \sqrt{4\left( 3\cdot 10^{k-1} + \frac{8}{9}\cdot 10^{k-1} - \frac{80}{9} + 9\right) } = \sqrt{4\left( \frac{35}{9} \cdot 10^{k-1} + \frac{1}{9}\right)} = \sqrt{\frac{140}{9}\cdot 10^{k-1} + \frac{4}{9} } = \sqrt{\frac{14}{9} \cdot 10^k + 1 - \frac{5}{9} }= \sqrt{1 + 10^k + \frac{5}{9}\cdot 10^k - \frac{5}{9} } $$ and the latter is the expression on the left.