User talk:Harshit 97

IIT 2009 question
Prove That the numbers 49, 4489, 444889, .... obtained by insering 48 into the middle of the preceding number are squares of integers.

444...n times x 10^n + 8888...n times + 1                                                      , 6666...n times[(2x10^n)/3+4/3] +1       , 666...n times[0.666...upto ∞ x 10^n + 1.333... upto∞]+1                                         , 6666...ntimes[666...upto n terms+(0.666... upto ∞ +1.333...∞)] +1                                , 6666...ntimes[6666...upto n times + (2)]+1                                                        , let 6666... ntimes be A                                                                            , A[A+2] +1                                                                                           , A^2 + 2A +1                                                                                          , (A+1)^2                                                                                               , therefore the no. obtained will be a whole square                                                      , and A+1 always an integer for any integral value of n                                                   , --Harshit 97 (talk) 06:59, 4 July 2012 (UTC)