<s>[INST] 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. 
 [/INST] 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.
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