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

Answer: (C)

$2^n-1$

Given that $f(n+1)=2f\left(n\right)+1,n\ge 1$.

Therefore, $f\left(2\right)=2f\left(1\right)+1$
Since $f\left(1\right)=1$, we have
$f\left(2\right)=2f\left(1\right)+1=2\left(1\right)+1=3=2^2-1$.
Similarly $f\left(3\right)=2f\left(2\right)+1=2\left(3\right)+1=7=2^3-1$

and so on....

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