Question ID: #771
Let $f$ and $g$ be functions satisfying $f(x+y)=f(x)f(y)$, $f(1)=7$ and $g(x+y)=g(xy)$, $g(1)=1$, for all $x, y\in\mathbb{N}$. If $\sum_{x=1}^{n}(\frac{f(x)}{g(x)})=19607$, then n is equal to:
- (1) 7
- (2) 5
- (3) 6
- (4) 4
Solution:
**Determine f(x)**
Given $f(x+y) = f(x)f(y)$. This implies $f(x) = a^x$.
Since $f(1) = 7$, we have $a^1 = 7 \Rightarrow a = 7$.
$$ f(x) = 7^x $$
**Determine g(x)**
Given $g(x+y) = g(xy)$.
Put $y=1$: $g(x+1) = g(x)$.
This means $g(x)$ is a constant function for $x \in \mathbb{N}$.
Since $g(1) = 1$, $g(x) = 1$ for all $x \in \mathbb{N}$.
**Solve the Summation**
$$ \sum_{x=1}^{n} \frac{f(x)}{g(x)} = \sum_{x=1}^{n} \frac{7^x}{1} = 7^1 + 7^2 + \dots + 7^n $$
This is a Geometric Progression with $a=7, r=7$.
$$ S_n = \frac{7(7^n – 1)}{7 – 1} = \frac{7}{6}(7^n – 1) $$
Given Sum = 19607:
$$ \frac{7}{6}(7^n – 1) = 19607 $$
$$ 7^n – 1 = \frac{19607 \times 6}{7} $$
$$ 7^n – 1 = 2801 \times 6 $$
$$ 7^n – 1 = 16806 $$
$$ 7^n = 16807 $$
We know powers of 7:
$7^1=7, 7^2=49, 7^3=343, 7^4=2401, 7^5=16807$.
$$ n = 5 $$
Ans. (2)
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