prove that (111....1)n times(55...5)(n-1) times 6 is always a perfect square.?

Answer 1

   1..{n}..15..{n-1}..56
= 1..{n}..10..{n}..0 + 5..{n}..5 + 1
= (10^n-1)/9 10^n + 5 (10^n - 1)/9 + 1
= [ (10^n - 1) 10^n + 5 (10^n - 1) + 9 ] / 9
= [ 10^(2n) + 4 10^n + 4 ] / 9
= [ (10^n + 2) / 3 ]²
= [ 3..{n-1}..3 4 ]²

Example, for n=4

     11115556 = 3334²

Answer 2

Hrm, unless I misunderstand the question, this statement is false.

Even the simplest case fails:

If n = 2, then you have 11 x 5 x 6 = 330.

330 is not a perfect square.

Are you sure you posted it correctly?