s=(3/1!+2!+3!)+(4/2!+3!+4!)+.............+(2008/2006!+2007!+2008!)then s is :A)1/2 - 1/2006!B)1/2 - 1/2008!c)1/2006! - 1/2008!D)1/2007! - 1/2008!

Answer is option b).

The 'r' th term can be written as (r+2)/r!+(r+1)!+(r+2)! = (r+1)/(r+2)! = 1/(r+1)! - 1/(r+2)!.

This sum can be telescoped(finite difference method) to get the required answer.