If `f(x)=(a^(x))/(a^(x)+sqrt(a))(agt0),g(n)=sum_(r=1)^(2n-1)2f((r)/(2n))`. Find te value `g(4)`

Explanation

To solve the problem, we need to evaluate the function \( g(n) \) defined as:

\[ g(n) = \sum_{r=1}^{2n-1} 2f\left(\frac{r}{2n}\right) \]

where

\[ f(x) = \frac{a^x}{a^x + \sqrt{a}} \quad (a > 0) \]

We are specifically asked to find \( g(4) \).

### Step 1: Calculate \( g(4) \)

Substituting \( n = 4 \) into the equation for \( g(n) \):

\[ g(4) = \sum_{r=1}^{2(4)-1} 2f\left(\frac{r}{8}\right) = \sum_{r=1}^{7} 2f\left(\frac{r}{8}\right) \]

### Step 2: Evaluate \( f\left(\frac{r}{8}\right) \)

Now we need to evaluate \( f\left(\frac{r}{8}\right) \):

\[ f\left(\frac{r}{8}\right) = \frac{a^{\frac{r}{8}}}{a^{\frac{r}{8}} + \sqrt{a}} \]

### Step 3: Simplify \( f\left(\frac{r}{8}\right) \)

We can express \( \sqrt{a} \) as \( a^{\frac{1}{2}} \):

\[ f\left(\frac{r}{8}\right) = \frac{a^{\frac{r}{8}}}{a^{\frac{r}{8}} + a^{\frac{1}{2}}} \]

### Step 4: Rewrite the sum

Now substituting \( f\left(\frac{r}{8}\right) \) back into the sum:

\[ g(4) = \sum_{r=1}^{7} 2 \cdot \frac{a^{\frac{r}{8}}}{a^{\frac{r}{8}} + a^{\frac{1}{2}}} \]

### Step 5: Factor out common terms

Factor out \( a^{\frac{1}{2}} \) from the denominator:

\[ g(4) = \sum_{r=1}^{7} 2 \cdot \frac{a^{\frac{r}{8}}}{a^{\frac{1}{2}}(a^{\frac{r}{8 - \frac{1}{2}} + 1)}} \]

This simplifies to:

\[ g(4) = \frac{2}{\sqrt{a}} \sum_{r=1}^{7} \frac{a^{\frac{r}{8}}}{a^{\frac{r}{8}} + 1} \]

### Step 6: Evaluate the sum

Now, we can evaluate the sum \( \sum_{r=1}^{7} \frac{a^{\frac{r}{8}}}{a^{\frac{r}{8}} + 1} \). Notice that:

\[ \frac{a^{\frac{r}{8}}}{a^{\frac{r}{8}} + 1} + \frac{1}{a^{\frac{r}{8}} + 1} = 1 \]

Thus, we can pair terms in the sum:

\[ g(4) = \frac{2}{\sqrt{a}} \cdot 7 \cdot \frac{1}{2} = \frac{7}{\sqrt{a}} \]

### Final Step: Conclusion

Thus, the final result for \( g(4) \) is:

\[ g(4) = \frac{7}{\sqrt{a}} \]