Hi Sathyaram,

Let 2011 = x

So x(x+1)(x+2)(x+3) + 1 = (x²+3x)(x²+3x+2) + 1 = (x²+3x)² + 2(x²+3x) + 1 = (x²+3x+1)²

So square root of that would be x²+3x+1

Where x = 2011.

Which is 2001²+3*2011+1 = 20011*2014+1 = 4030015.

Hope that helps.

All the best.

Regards,

Ashwin (IIT Madras).

wel...brother...i think this question appeared in F-NSTSE. Thought it looks complicated but its not that hard..lets see :

let 2011=x, Then 2012=(x+1), 2013=(x+2) & 2014=(x+3)

Then we have:

M = √x(x+1)(x+2)(x+3)+1            .... (say)

   =√[x(x+3)][(x+1)(x+2)]+1

  = √[(x²+3x)(x²+3x+2)]+1

  = √[(x²+3x+1)-1][(x²+3x+1)+1]+1

  = √[(x²+3x+1)²-1²]+1

  = √(x²+3x+1)²-1+1

  =  √(x²+3x+1)²

  =  x²+3x+1 

Therefore, M = 2011² + 3(2011) +1 = 4,050,155

I hope this help u...:) 

I am studying in 10th standard.I want to talk to you through this website,how can i talk to you privately?Eager to know your reply.

Regards,

V.Sathyaram