If f(x) is a function satisfying f(x + y) = f | vedclass

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(A) Given that $f(x + y) = f(x)f(y)$ for all $x, y \in N$.For $x = 1$,$f(2) = f(1 + 1) = f(1)f(1) = 3^2 = 9$.For $x = 2$,$f(3) = f(2 + 1) = f(2)f(1) =

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$4$

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(A) Given that $f(x + y) = f(x)f(y)$ for all $x, y \in N$.For $x = 1$,$f(2) = f(1 + 1) = f(1)f(1) = 3^2 = 9$.For $x = 2$,$f(3) = f(2 + 1) = f(2)f(1) = 3^2 \cdot 3 = 3^3 = 27$.By induction,$f(x) = 3^x$.Now,we are given $\sum_{x = 1}^n f(x) = 120$.This is a geometric series: $3^1 + 3^2 + 3^3 + \dots + 3^n = 120$.The sum of a geometric series is given by $S_n = \frac{a(r^n - 1)}{r - 1}$,where $a = 3$ and $r = 3$.So,$\frac{3(3^n - 1)}{3 - 1} = 120$.$\frac{3(3^n - 1)}{2} = 120$.$3(3^n - 1) = 240$.$3^n - 1 = 80$.$3^n = 81$.$3^n = 3^4$.Therefore,$n = 4$.

If $f(x)$ is a function satisfying $f(x + y) = f(x)f(y)$ for all $x, y \in N$ such that $f(1) = 3$ and $\sum_{x = 1}^n f(x) = 120$,then the value of $n$ is

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