Question

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

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

Answer 1

The correct option is C $2^n-1$

$f\left(1\right)=1$ $n\ge 1$

$f(n+1)=2f\left(n\right)+1f\left(2\right)=2f\left(1\right)+13=2^2-1f\left(3\right)=2.3+17=2^3-1f\left(4\right)=2.7+1=15=2^4-1$

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

Answer C