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f(1)=1, n≥1⇒f(n+1)=2f(n)+1 then f(n) =

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a

2n+1

b

2n

c

2n−1

d

2n−1−1

answer is C.

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Detailed Solution

Put n = 1f(2)=3⇒f(2)=22-1 f(3)=7⇒f(3)=23-1..... f(n)=2n-1

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f(1)=1, n≥1⇒f(n+1)=2f(n)+1 then f(n) =