Question

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

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

Answer 1

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

$f(n+1)=2f\left(n\right)+1\to 1$
Replace $n$ by $n-1$
$\Rightarrow f\left(n\right)=2f(n-1)+1\to 2$
Equating both equation , we get
$f(n+1)=4f(n-1)+1$
$=2.2f(n-1)+1$
$\therefore f\left(n\right)=2f(n-1)$ and so on
$f(n-1)=2f(n-1)$
$\therefore f\left(n\right)=2^{n}f\left(0\right)+9$
$f\left(1\right)=1+9=1$
$\Rightarrow c=-1$
$\therefore f\left(n\right)=2^n-1$