Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square 
Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square. 

We can observe that $49=7^2, 4489=67^2, 444889=667^2, \ldots$
I have tried expanding terms of the sequence, and to express it as a whole square. But it was too tough for me. 
Any help will be appreciated. 
 A: The $n$th term is $T_{n}=9+8.10+8.10^2+...8.10^{n-1}+4.10^n+...+4.10^{2n-1}$:
$$
\begin{align}
T_n &= 9+8.10(1+10+...10^{n-2})+4.10^n(1+...+10^{n-1}) \\
    &= 9+8.10(\frac{10^{n-1}-1}{9})+4.10^n (\frac{10^{n}-1}{9}) \\
    &=9+8.\frac{10^{n}-10}{9}+4.\frac{10^{2n}-10^n}{9} \\
    &=\frac{81+8.10^n-80+4.10^{2n}-4.10^n}{9} \\
    &=\frac{1}{9}.(1+4.10^n+4.10^{2n}) \\
    &={(\frac{{1+2.10^n}}{3}})^2
\end{align}
$$.
Now $10\equiv 1\pmod{3}\Rightarrow 10^n\equiv 1\pmod{3}\Rightarrow 2.10^n+1\equiv 0\pmod{3}$, thus $3$ divides the numerator and we are done.
A: Very similar to this question, which asks for a proof that $1111\ldots5556$ is a square. The proof is just as simple:
Multiply $4444\ldots8889$ by $9$ and you get $4000\ldots0004000\ldots0001$, which is $2000\ldots0001^2$.
A: We can prove it as follows:
\begin{align*}
\underbrace{666\ldots6}_{n}\,7^2
&= \left[7 + 6(10^1 + 10^2 + \cdots + 10^n) \right]^2 \\
&= \left[7 + 6 \frac{10^{n+1} - 10}{9}\right]^2 \\
&= \left[ \frac{2 \cdot 10^{n+1} + 1}{3}\right]^2 \\
&= \frac{4 \cdot 10^{2n+2} + 4 \cdot 10^{n+1} + 1}{9} \\
&= 4 \cdot \frac{10^{2n+2} - 1}{9} + 4 \cdot \frac{10^{n+1} - 1}{9} + 1 \\
&= 4 \cdot \underbrace{111 \ldots 1}_{2n+2} + 4 \cdot \underbrace{111 \ldots 1}_{n+1} + 1 \\
&= \underbrace{444 \ldots 4}_{2n+2} + \underbrace{444 \ldots 4}_{n+1} + 1 \\
&= \underbrace{444 \ldots 4}_{n+1}\,\underbrace{888 \ldots 8}_{n}\,9.
\end{align*}
A: You can see that $$(1+\sum_{i=0}^{n+1}6\times10^i)^2=(6\times10^{n+1}+(1+\sum_{i=0}^{n}6\times10^i))^2.$$ Expand it to get the expression $$(6\times10^{n+1})^2+2\times6\times10^{n+1}\times(1+\sum_{i=0}^{n}6\times10^i)+(1+\sum_{i=0}^{n}6\times10^i)^2.$$ Now make an induction hypothesis that the squares of $66...67$ makes the sequence in the question. Induction basis is already proven. Then the last term in the expanded expression becomes $$\color{red}{\sum_{i=n+1}^{2n+1}4\times10^i+\sum_{j=1}^{n}8\times10^j+9}.$$ The second term in the expanded expression can be simplified by taking the factor $3$ of $6$ and multiply it with $(1+\sum_{i=0}^{n}6\times10^i)$, giving a simple expression $2\times10^{n+1}+1.$ With this expression, the second term becomes $$2\times2\times10^{n+1}\times(2\times10^{n+1}+1)=\color{red}{8\times10^{2n+2}+4\times10^{n+1}}.$$ The first term is just $\color{red}{36\times10^{2n+2}}$. You can then easily verify that the sum of the three terms (in red) is the expected expression.
