Ramaiah entrance test question [Solution]

The problem was to show that  the sequence of numbers

$$49, 4489, 444889, \dots$$

where each number is formed by inserting 48 in the middle of the previous number:

$$49 \to 4\underline{48}9 \to 44\underline{48}89 \to \dots$$

has only perfect squares!

Solution

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Any number is of the form $n$ 4s followed by $n-1$ 8s followed by 9.

This can be rewritten as the sum of

$4444\dots44$ (a $2n$ digit number) 

$44\dots4$  (an $n$ digit number)

1

This is basically

$$\frac{4(10^{2n}-1)}{9} + \frac{4(10^{n}-1)}{9}  + 1$$

$$= \left(\frac{2\times10^n + 1}{3}\right)^2$$