Let f(x) be a function such that f(x+y)=f(x) | vedclass

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(C) Given $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in \mathbb{N}$ and $f(1)=3$.We can find the terms of the sequence:$f(1)=3$$f(2)=f(1+1)=f(1) \cdot f(

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$7$

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(C) Given $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in \mathbb{N}$ and $f(1)=3$.We can find the terms of the sequence:$f(1)=3$$f(2)=f(1+1)=f(1) \cdot f(1)=3^2=9$$f(3)=f(2+1)=f(2) \cdot f(1)=3^2 \cdot 3=3^3=27$In general,$f(k)=3^k$.The sum is given by $\sum_{k=1}^{n} f(k) = \sum_{k=1}^{n} 3^k = 3279$.This is a geometric progression with first term $a=3$,common ratio $r=3$,and $n$ terms.The sum formula is $S_n = \frac{a(r^n-1)}{r-1}$.Substituting the values:$\frac{3(3^n-1)}{3-1} = 3279$$\frac{3(3^n-1)}{2} = 3279$$3(3^n-1) = 6558$$3^n-1 = 2186$$3^n = 2187$Since $3^7 = 2187$,we have $n=7$.

Let $f(x)$ be a function such that $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in \mathbb{N}$. If $f(1)=3$ and $\sum_{k=1}^{n} f(k)=3279$,then the value of $n$ is:

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