Q.
If f(1)=1, f(n+1)=2f(n)+1 and n≥1,then f(n) is equal to
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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
Given ,f(1)=f(n+1)=2f(n)+1,n≥1 f(2)=f(1+1)=2f(1)+1=2*1+1=3=22-1 f(3)=f(2+1)=2f(2)+1=2*3+1=7=23-1 f(4)=f(3+1)=2f(3)+1=2*7+1=15=24-1 f(5)=f(4+1)=2f(4)+1=2*15+1=31=25-1f(n)=f((n-1)+1)=2f(n-1)+1=2n-1
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