Perfect squares

$49, 4489, 444889, 44448889.................$ In the series of the above numbers following the same pattern, prove that, each of these are whole squares of integers..

#NumberTheory #PerfectSquare #sequence

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

$T_n = 44444....(n times) 88888... (n-1 times)9$

$= \frac{4}{9}\times(10^n-1) \times 10^n + \frac{8}{9} \times(10^{n-1}-1) \times 10 + 9$

$= \frac{1}{9}({ 4\times(10^n-1)\times10^n + 8(10^{n-1}-1)\times + 81})$

$= \frac{1}{9}{( 4 \times 10^{2n} - 4 \times 10^n + 8 \times(10^n - 10) + 81)}$

$= \frac{1}{9}{( 4\times10^{2n} - 4\times10^n + 8 \times 10^n - 80 + 81)}$

$= \frac{1}{9}{( 4\times10^{2n} + 4 \times 10^n + 1)}$

$= \frac{1}{9}[(2\times10^n)^2 + 2\times (2\times10^n) + 1^2]$

$= \frac{1}{9}(2\times10^n + 1)^2$

$= (\dfrac{2 \times 10^n +1}{3})^2$

Now note

$2 \equiv -1 (\mod{3} )$ and

$10 \equiv 1 (\mod{3} ) \implies 10^n \equiv 1 (\mod{3} )$

$\implies 2\times10^n \equiv -1 (\mod{3} )$

$\implies 2\times10^n + 1 \equiv 0 (\mod{3} )$ or

$3 | 2\times10^n + 1$ .

This maybe proved by induction as well. Thus ${(2*10^n + 1)/3}$ is an integer and thus $T_n$ is a perfect square for all n.

Sagnik Saha - 7 years, 1 month ago

:) 9801,998001,99980001,9999800001,.....

Ritabrata Roy - 2 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$