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

Note by Sagnik Saha 4 years, 1 month ago

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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.

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