Question

if f(1) = 1 , f(n+1) = 2f(n) + 1 ,n >= 1 find f(n)

Answer 1

Answer:

$\bf\,f(n)=2^n-1$

Step-by-step explanation:

$\textbf{Given:}$

$f(n+1)=2[f(n)+1]$

$=2[2f(n-1)+1]+1$

$=2^2f(n-1)+2+1$

$=2^2[2f(n-2)+1]+2+1$

$=2^3f(n-2)+2^2+2+1$

$\text{Proceeding like this, we get}$

$f(n+1)=2^n\,f(n-(n-1))+2^{n-1}+2^{n-2}+.....+1$

$=2^n\,f(1)+2^{n-1}+....+1$

$=2^n(1)+2^{n-1}+2^{n-2}+...2+1$

$=1+2+2^2+...2^n$

$\text{This is a G.P with a=1 and r=2 containing n+1 terms}$

$=\frac{1(2^{n+1}-1)}{2-1}$

$f(n+1)=2^{n+1}-1$

$\implies\,2f(n)+1=2^{n+1}-1$

$\implies\,2f(n)=2^{n+1}-2$

$\implies\,\boxed{\bf\,f(n)=2^n-1}$