If (x= 2 5 + 1 3 2 ! ( 2 5 )^2+ 1 3 5 3 ! ( 2 5 )^3+ ), then (x+ 1 x = )

Step-by-step solution

Let (x=(1+y)^n-1=n y ) ( aligned + & n(n-1) 2 ! y^2+ n(n-1)(n-2) 3 ! y^3+ . . & = 2 5 + 1.3 2 ! ( 2 5 )^2+ 1 3 5 3 ! ( 2 5 )^3+ . . . aligned ) On comparing first three terms, we get (n y= 2 5 , n(n-1) 2 ! y^2= 1 3 2 ! ( 2 5 )^2 ) and ( n(n-1)(n-2) 3 ! y^3= 1 3 5 3 ! ( 2 5 )^3 ) From first two relations, we get ( array rlrl & & n y(n y-y) 2 ! & = 2 5 ( 2 5 -y ) 2 ! = 1 3 2 ! ( 2 5 )^2 & ( 2 5 )-y & =1 3 ( 2 5 ) & & y & = 2 5 - 6 5 =- 4 5 and n=- 1 2 array ) Now, on putting (n=- 1 2 ) and (y=- 4 5 ) in LHS of third