<s>[INST] Square of $7+\sum_{k=1}^n6\times10^k$ If we build a number as follow: 
$$N=7+\sum_{k=1}^n6\times10^k$$
we find:
$$N^2=9+\sum_{k=1}^n8\times10^k+\sum_{j=n+1}^{2n+1
}4\times10^{j}$$
that means for example: 
$67^2=4489$,
$667^2=444889$,
$6667^2=44448889$
and so on. How can we prove that given an arbitrary $n$ and a number $N$ built as above, the square of $N$ can be obtained by th equation for $N^2$? Thanks
 [/INST] We have 
$$N=1+6\sum_{k=0}^n10^k=1+6\cdot \frac{10^{n+1}-1}9=\frac{2\cdot 10^{n+1}-1}3.$$
Hence 
$$N^2=\frac{4\cdot 10^{2n+2}-4\cdot 2^{n+1}+1}{9}=4\cdot 10^{n+1}\cdot \frac{10^{n+1}-1}{9} +8\cdot \frac{10^{n+1}-1}{9}+1$$
</s>