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

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(A) Given that $f(x + y) = f(x) f(y)$ for all $x, y \in N$.By induction,$f(x) = [f(1)]^x$.Since $f(1) = 3$,we have $f(x) = 3^x$.Now,we are given $\sum

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$4$

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(A) Given that $f(x + y) = f(x) f(y)$ for all $x, y \in N$.By induction,$f(x) = [f(1)]^x$.Since $f(1) = 3$,we have $f(x) = 3^x$.Now,we are given $\sum_{x=1}^n f(x) = 120$.Substituting $f(x) = 3^x$,we get $\sum_{x=1}^n 3^x = 120$.This is a geometric series with first term $a = 3$,common ratio $r = 3$,and $n$ terms.The sum is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Substituting the values: $\frac{3(3^n - 1)}{3 - 1} = 120$.$\frac{3(3^n - 1)}{2} = 120$.$3(3^n - 1) = 240$.$3^n - 1 = 80$.$3^n = 81$.$3^n = 3^4$.Therefore,$n = 4$.

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

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