$49,4489,444889,44448889$ is a perfect square. [duplicate]
Show that each number in the sequence $49,4489,444889,44448889$ is a perfect square.
My attempt is basically a common one.
$a_4=4444(10^4)+888(10)+9$
$=4(1111)(10^4)+8(111)(10)+9$
$=4(\frac{10^4-1}{9})(10^4)+8(\frac{10^3-1}{9})(10)+9$
$a_n=(\frac{10^n-1}{9})(10^n)+8(\frac{10^{n-1}-1}{9})(10)+9$
After some indices work, I managed to get $a_n=(\frac{2(10^n)+1}{3})^2$, which shows that each number in this sequence is a perfect square.
I'm basically has zero knowledge about number theory. Does any another way to solve this question? Thanks in advance.
$\endgroup$ 11 Answer
$\begingroup$Here's another approach.
Usually when you see repeated numbers, try multiplying by $9$ to remove the repeats.
Doing this gives $$441, 40401, 4004001, 400040001,\cdots$$ which is the series $$4(10^{2n})+4(10^n)+1=(2(10^n)+1)^2$$
i.e. a perfect square.
Since $9$ is also a perfect square, the original series is also made up of perfect squares.
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