Question

If $f(1)=1$ and $f(n+1)=2f(n)+1$ if $n\ge 1$, then $f(n)$ is equal to

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

Answer 1

Answer: (C)

$f\left(1\right)=1$ and $f\left(n+1\right)=2f\left(n\right)+1,n\ge 1$.
$\therefore f\left(2\right)=2\left(1\right)+1=3,f\left(3\right)=7,f\left(4\right)=15,\dots$ and so on
Thus, $f\left(1\right)=2^1-1,f\left(2\right)=2^2-1$,
$f\left(3\right)=2^3-1,\dots ,f\left(n\right)$
$=2^n-1$.