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(B) Given the functional equation $f(x+y) = f(x)f(y)$.For $x=1, y=1$,we have $f(2) = f(1)f(1) = 3^2 = 9$.For $x=2, y=1$,we have $f(3) = f(2)f(1) = 3^2

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$5$

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(B) Given the functional equation $f(x+y) = f(x)f(y)$.For $x=1, y=1$,we have $f(2) = f(1)f(1) = 3^2 = 9$.For $x=2, y=1$,we have $f(3) = f(2)f(1) = 3^2 \cdot 3 = 3^3 = 27$.By induction,it follows that $f(x) = 3^x$ for all $x \in R$.We are given the sum $\sum_{i=1}^{n} f(i) = 363$,which implies $\sum_{i=1}^{n} 3^i = 363$.This is a geometric progression with the first term $a = 3$,common ratio $r = 3$,and $n$ terms.The sum of a geometric series is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Substituting the values: $\frac{3(3^n - 1)}{3 - 1} = 363$.$\frac{3(3^n - 1)}{2} = 363$.$3(3^n - 1) = 726$.$3^n - 1 = 242$.$3^n = 243$.Since $243 = 3^5$,we have $n = 5$.

Suppose that a function $f: R \rightarrow R$ satisfies $f(x+y)=f(x) f(y)$ for all $x, y \in R$ and $f(1)=3$. If $\sum_{i=1}^{n} f(i)=363$,then $n$ is equal to

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