f(1)=1, n≥1⇒f(n+1)=2f(n)+1 then f(n) =
2n+1
2n
2n−1
2n−1−1
Put n = 1
f(2)=3⇒f(2)=22-1 f(3)=7⇒f(3)=23-1..... f(n)=2n-1