Question

If $f:N\to N$, where $N$ is set of natural numbers and $f\left(xy\right)=f\left(x\right)\cdot f\left(y\right)-f(x+y)+1\forall x,y\in N$ and $f\left(1\right)=2$, then the value of $1+\sum _{k=1}^{\infty }\frac{f\left(k\right)}{2^k}$ is

Answer 1

Given : $f\left(xy\right)=f\left(x\right)\cdot f\left(y\right)-f(x+y)+1$
Putting $y=1$, we get
$f\left(x\right)=2f\left(x\right)-f(x+1)+1\Rightarrow f(x+1)-f\left(x\right)=1f\left(2\right)-f\left(1\right)=1f\left(3\right)-f\left(2\right)=1f\left(4\right)-f\left(3\right)=1⋮f(x+1)-f\left(x\right)=1\Rightarrow f(x+1)-f\left(1\right)=x\Rightarrow f(x+1)=x+2\therefore f\left(x\right)=x+1$
Now, let
$S=\sum _{k=1}^{\infty }\frac{f\left(k\right)}{2^k}\Rightarrow S=\sum _{k=1}^{\infty }\frac{x+1}{2^k}\Rightarrow S=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}\cdots \cdots \Rightarrow \frac{S}{2}=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots \cdots \Rightarrow \frac{S}{2}=1+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots \cdots \Rightarrow \frac{S}{2}=1+\frac{\frac{1}{4}}{1-\frac{1}{2}}\Rightarrow S=3\therefore 1+\sum _{k=1}^{\infty }\frac{f\left(k\right)}{2^k}=1+3=4$