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 displaystyle∑k=1n f(k)=3279, then the value of n is

Question

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 displaystyle∑k=1n f(k)=3279, then the value of n is (A) 8 (B) 6 (C) 7 (D) 9

Tardigrade Admin · Accepted Answer

f ( x + y )= f ( x ) ⋅ f ( y ) ∀ x , y ∈ N , f (1)=3 f (2)= f 2(1)=32 f (3)= f (1) f (2)=33 f (4)=34 f ( k )=3 k displaystyle ∑ k =1 n f ( k )=3279 f (1)+ f (2)+ f (3)+ ldots ldots ldots+ f ( k )=3279 3+32+33+ ldots ldots ldots 3 k =3279 (3(3 k -1)/3-1)=3279 (3 k -1/2)=1093 3 k -1=2186 3 k =2187 k =7

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=1∑n​f(k)=3279, then the value of n is

8

6

7

9

f(x+y)=f(x)⋅f(y)∀x,y∈N,f(1)=3 f(2)=f2(1)=32 f(3)=f(1)f(2)=33 f(4)=34 f(k)=3k k=1∑n​f(k)=3279 f(1)+f(2)+f(3)+………+f(k)=3279 3+32+33+………3k=3279 3−13(3k−1)​=3279 23k−1​=1093 3k−1=2186 3k=2187 k=7

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 $k = 1 \sum n f (k) = 3279$ , then the value of $n$ is