$$\frac{f\left(2\right)}{f\left(1\right)}=>\ This\ \exp ression\ looks\ like\ the\ common\ ratio\ of\ a\ geometric\ sequence.$$
$$If\ first\ term=\frac{f\left(2\right)}{f\left(1\right)};$$
$$Second\ term=\frac{f\left(3\right)}{f\left(2\right)};$$
$$nth\ term=\frac{f\left(n+1\right)}{f\left(n\right)};$$
Then this could be an arithmetic sequence as well. We need to find the value of at least 2 unknown functions to decipher if we will finding the sum of AP or GP.
Statement 1=> f(1)=1
$$f\left(x\right)=x\ or\ f\left(x\right)=x^0\ or\ f\left(x\right)=\left(x+1\right)-1$$
No expression is given to validate this statement if we define any expression that's true for x=1, we might get varying results when x=2. Hence, statement 1 is NOT SUFFICIENT.
Statement 2=> f(g+b) = f(g) * f(b)
i.e f(3) = f(2+1) = f(2) * f(1)
and f(2) = f(1+1) = f(1) * f(1).
This statement tells us that the sum of all the arguments in a function can be expressed as product of a function of its own. But to solve the question with this statement, we need to know the exact value of f(1). And since there is no expression to solve for f(1), statement 2 is NOT SUFFICIENT.
Combining both statements together/u]
S1: f(1)=1
S2: f(g+b) = f(g) * f(b)
$$Therefore,\ f\left(2\right)=f\left(1+1\right)=f\left(1\right)\cdot f\left(1\right)=1\cdot1=1$$ $$f\left(3\right)=f\left(2+1\right)=f\left(2\right)\cdot f\left(1\right)=1\cdot1=1$$
$$f\left(4\right)=f\left(2+2\right)=f\left(2\right)\cdot f\left(2\right)=1\cdot1=1$$
$$f\left(n\right)=1$$
$$So\ therefore,\ f\left(1\right),\ f\left(2\right),\ f\left(3\right),...,f\left(2006\right)\ all\ have\ the\ same\ values\ which\ is\ 1.$$
$$The\ value\ of\ \frac{f\left(2\right)}{f\left(1\right)}=\frac{1}{1}=1$$
$$\frac{f\left(3\right)}{f\left(2\right)}=\frac{1}{1}=1$$
$$With\ careful\ observation\ from\ \frac{f\left(2\right)}{f\left(1\right)}to\frac{f\left(5\right)}{f\left(4\right)},$$
$$\frac{f\left(2\right)}{f\left(1\right)}+\frac{f\left(3\right)}{f\left(2\right)}+\frac{f\left(4\right)}{f\left(3\right)}+\frac{f\left(5\right)}{f\left(4\right)}=1+1+1+1=4$$
$$Sum\ of\ \frac{f\left(2\right)}{f\left(1\right)}to\ \frac{f\left(10\right)}{f\left(9\right)}=>$$
$$\frac{f\left(2\right)}{f\left(1\right)}+\frac{f\left(3\right)}{f\left(2\right)}+\frac{f\left(4\right)}{f\left(3\right)}+\frac{f\left(5\right)}{f\left(4\right)}+\frac{f\left(6\right)}{f\left(5\right)}+\frac{f\left(7\right)}{f\left(6\right)}+\frac{f\left(8\right)}{f\left(7\right)}+\frac{f\left(9\right)}{f\left(8\right)}+\frac{f\left(10\right)}{f\left(9\right)}=$$
$$1+1+1+1+1+1+1+1+1=9$$
$$Sum\ of\ \frac{f\left(2\right)}{f\left(1\right)}to\ \frac{f\left(n+1\right)}{f\left(n\right)}=>n$$
$$i.e\ \frac{f\left(2\right)}{f\left(1\right)}+\frac{f\left(3\right)}{f\left(2\right)}+...+\ \frac{f\left(n+1\right)}{f\left(n\right)}=>n$$
If n=4, we have;
$$\frac{f\left(2\right)}{f\left(1\right)}+\frac{f\left(3\right)}{f\left(2\right)}+\frac{f\left(4\right)}{f\left(3\right)}+\frac{f\left(5\right)}{f\left(4\right)}=4$$
If n=2005, we have;
$$\frac{f\left(2\right)}{f\left(1\right)}+\frac{f\left(3\right)}{f\left(2\right)}+\frac{f\left(4\right)}{f\left(3\right)}+\frac{f\left(5\right)}{f\left(4\right)}+\frac{f\left(2006\right)}{f\left(2005\right)}=2005$$.
Conclsuively, both statements combined together ARE SUFFICIENT.
Answer = option C
Thanks<i class="em em-v"></i>
