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Question

If $$ f(1) = 1, f (n+1) = 2 f (n) + 1 , n \ge 1 , $$ then $$ f(n) $$ is :


A
2n+1
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B
2n
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C
2n1
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D
2n11
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Solution

The correct option is C $$2^{n} -1$$
In mathematics, a real-valued function or real function is a function whose values are real numbers. In other words, it is a function that assigns a real number to each member of its domain.
Given,  $$f(1)=1$$  and $$f(n+1)=2f(n)+1,\quad n\ge 1$$
For $$n=1$$  $$\rightarrow $$ $$f(2)=2f(1)+1=3={ 2 }^{ 2 }-1$$
For $$n=2$$ $$\rightarrow $$ $$f(3)=2f(2)+1=3={ 2 }^{ 3 }-1$$
For $$n=3$$ $$\rightarrow $$ $$f(4)=2f(3)+1=3={ 2 }^{ 4 }-1$$
For $$n=4$$ $$\rightarrow $$ $$f(5)=2f(4)+1=3={ 2 }^{ 5 }-1$$
As seen from the above value pattern, 
$$f(n)={ 2 }^{ n }-1$$

Mathematics

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