Let f ( x ) be a function such that f ( x + y ) = f ( x ) · f ( y ) for all x , y N . If f ( 1 ) = 3 and k = 1 n f ( k ) = 3279 , then the value of n is [2023]

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Solution

Q.
Let $f (x)$ be a function such that $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in N .$ If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) = 3279,$ then the value of n is [2023]

1 8

2 6

3 7

4 9

Ans.
(3)

$f (x + y) = f (x) \cdot f (y) \forall x, y \in N$

Put $x = y = 1, f (2) = 3^{2}; Put x = 2, y = 1$

$f (3) = 3^{3}, f (4) = 3^{4}$ and so on.

Now, $\sum_{k = 1}^{n} f (k) = 3279 \Rightarrow 3 + 3^{2} + 3^{3} + \dots + 3^{n} = 3279$

$\Rightarrow \frac{3 (3^{n} - 1)}{3 - 1} = 3279 \Rightarrow 3^{n} = 2187$ $∴$ $n = 7$

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