If `f(x+y+1)={sqrt(f(x))+sqrt(f(y))}^2` and `f(0)=1AAx ,y in R , d e t e r m i n e f(n),n in Ndot`

Answer

`f(n)=(n+1)^(2)`

Explanation

Given `f(x+y+1)=(sqrt(f(x))+sqrt(f(y)))^(2)`
Putting `x=y=0,` we get
`f(1)=(sqrt(f(0))+sqrt(f(0)))^(2)=(1+1)^(2)=2^(2)`
Again putting `x=0,y=1` we get
`f(2)=(sqrt(f(0))+sqrt(f(1)))^(2)=(1+2)^(2)=3^(2)`
and for `x=1,y=1,` we get
`f(3)=(sqrt(f(1))+sqrt(f(1)))^(2)=(2+2)^(2)=4^(2)`
Hence, `f(n)=(n+1)^(2)`.