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]

Question

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]

Smart Achievers · Accepted Answer

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

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

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

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Solution

Q. 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=1nf(k)=3279, then the value of n is [2023]

1 8

2 6

3 7

4 9

Ans. (3) f(x+y)=f(x)·f(y) ∀x,y∈N Put x=y=1, f(2)=32; Put x=2,y=1 f(3)=33, f(4)=34 and so on. Now, ∑k=1nf(k)=3279⇒3+32+33+…+3n=3279 ⇒3(3n-1)3-1=3279⇒3n=2187 ∴ n=7

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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]