Square numbers sequence

(Facebook/Mathematics/BM – Problem by Laars Helenius)

Prove that each term of the following sequence is a perfect square:
49, 4489, 444889, 44448889, 4444488889, 444444888889, …
In general, the nth term has n 4’s and n-1 8’s and one 9

Proof:

Notation: <n> = 444…48…889 with n digits 4, etc (n>=1).
9*<n> = 10*<n> – <n> = 444…8…8890 – 444…4…8889
Here the first … in the terms mean 4:s, and the second … mean 8:s. The 8 in the first term and the 4 in the second one are in the same position (n digits to the right of them). Thus, 9*<n> = 4*102n + 4*10n + 1 = (2*10n + 1)2 where the last 1 comes from 90 – 89.

Now, 2*10n + 1 is obviously divisible by 3, i e 2*10n + 1 =3*m (m integer) so <n> = (3*m)2/9 = m2, i e <n> is a square. – qed

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