We have for $n^{th}$ number it is n 1s followed by n-1 5's and ending with 6
of it is n 1's follwed by n 5's and add 1
n '1s $\frac{1}{9}(10^n-1)$
n 5's is $\frac{5}{9}(10^n-1)$
so the number is $\frac{1}{9}(10^n-1) * 10^n + \frac{5}{9}(10^n-1) + 1$
$= \frac{1}{9}(10^{2n}-10^n + 5* (10^n-1) + 9)$
$= \frac{1}{9}(10^{2n} + 4* 10^n + 4)$
$=\frac{1}{9} (10^n + 2)^2$
$=(\frac{1}{3}(10^n+2))^2$
which is peffect square
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