First slide

Algebra of real valued functions

Question

If f(1)=1, f(n+1)=2f(n)+1 and n $\geq$ 1,then f(n) is equal to

Easy

A

.2ⁿ+1
B

2ⁿ
C

2ⁿ.-1
D

2^n-1-1

Solution

Given ,f(1)=f(n+1)=2f(n)+1,n $\geq$ 1
^{f(2)=f(1+1)=2f(1)+1=21+1=3=22-1}
^{f(3)=f(2+1)=2f(2)+1=23+1=7=23-1}
f(4)=f(3+1)=2f(3)+1=27+1=15=2⁴-1
f(5)=f(4+1)=2f(4)+1=215+1=31=2⁵-1
f(n)=f((n-1)+1)=2f(n-1)+1=2ⁿ-1

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