2016/094) Prove that in a sequence of numbers 49,4489, .... where each number is made by inserting 48 in the middle of previous number each number is a perfect square.

We have the $n^{th}$ number is n 4's followed by n 8's + 1
n 4's = $4 *(\frac{10^n-1}{9})$ n '8s = $8 *(\frac{10^n-1}{9})$
it may kindly be noted that the numerator  is divisible by 9
so the number = $4 *(\frac{10^-1}{9}) * 10^n + 8 *(\frac{10^-1}{9}) + 1$
$=\frac{4 * (10^n-1) * 10 ^n + 8(10^n-1) + 1}{9}$
$=\frac{ 4 * 10^{2n} + 4 * 10^n + 1}{9}$
$= \frac{(2 * 10^n)^2 + 2 * (2 * 20^n) + 1}{9}$
$= \frac{(2 * 10^n + 1)^2}{3^2}$
$= (\frac{2 * 10^n+1}{3})^2$
Beacuse numerator is divisible by 3 so it is a perfect square