Question
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 $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
The correct answer is
4
Function Properties Analysis
The function $f$ satisfies $f(x+y) = f(x)f(y)$ with $f(1) = 7$. This indicates an exponential function of the form $f(x) = a^x$.
Using $f(1)=7$, we find $a^1 = 7$, so $a=7$. Thus, $f(x) = 7^x$.
The function $g$ satisfies $g(x+y) = g(xy)$ with $g(1) = 1$. Let's determine $g(x)$:
- $g(2) = g(1+1) = g(1 \times 1) = g(1) = 1$.
- $g(3) = g(2+1) = g(2 \times 1) = g(2) = 1$.
- Continuing this pattern, $g(x) = 1$ for all $x \in \mathbb{N}$.
Summation Equation Solving
We need to solve the equation $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$.
Substitute $f(x) = 7^x$ and $g(x) = 1$:
$ \sum_{x=1}^{n} \frac{7^x}{1} = 19607 $
$ \sum_{x=1}^{n} 7^x = 19607 $
This is the sum of a geometric series with first term $a=7$, common ratio $r=7$, and $n$ terms.
Using the formula $S_n = a \frac{r^n - 1}{r-1}$:
$ S_n = 7 \frac{7^n - 1}{7-1} = 7 \frac{7^n - 1}{6} $
Set the sum equal to the given value:
$ 7 \frac{7^n - 1}{6} = 19607 $
Solve for $n$:
- Multiply by 6: $7(7^n - 1) = 19607 \times 6 = 117642$.
- Divide by 7: $7^n - 1 = \frac{117642}{7} = 16806$.
- Add 1: $7^n = 16806 + 1 = 16807$.
- Determine $n$: By testing powers of 7, we find $7^5 = 16807$.
Therefore, $n=5$.
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