IIT JEE , AIEEE Forum - Mathematics , Algebra

risin

Somebody

nadeemoidu

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This can be expressed as the sum of a GP with common ratio 10.

So ,

$a_n = \frac{10^{3^n}-1}{9}$

Similarly

$a_{n-1} = \frac{10^{3^{n-1}}-1}{9}$

Now ,

$a_n = \frac{10^{3^n} -1 }{9} \\=\frac{(10^{3^{n-1}})^3-1}{9}$

Expand this using the formula a^3 - b^3 . The first term will be a_(n-1) and it can be easily shown that the second term is divisible by3