Question

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

• A. $2^{n+1}$
• B. $2^n$
• C. $2^n-1$
• D. $2^{n-1}-1$

Answer 1

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, $f\left(1\right)=1$ and $f(n+1)=2f\left(n\right)+1,n\ge 1$

For $n=1$ $\to$ $f\left(2\right)=2f\left(1\right)+1=3=2^2-1$

For $n=2$ $\to$ $f\left(3\right)=2f\left(2\right)+1=3=2^3-1$

For $n=3$ $\to$ $f\left(4\right)=2f\left(3\right)+1=3=2^4-1$

For $n=4$ $\to$ $f\left(5\right)=2f\left(4\right)+1=3=2^5-1$

As seen from the above value pattern,

$f\left(n\right)=2^n-1$