The sum to infinity of the series
`1+(4)/(5)+(7)/(5^(2))+(10)/(5^(3))+ . . . ,` is

Options

A. `(16)/(35)`
B. `(11)/(8)`
C. `(35)/(16)`
D. `(8)/(6)`

Answer

To find the sum to infinity of the series \( S = 1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \ldots \), we can follow these steps:

### Step 1: Identify the series
We can observe that the series consists of two parts: the numerators \( 1, 4, 7, 10, \ldots \) and the denominators \( 1, 5, 5^2, 5^3, \ldots \).

### Step 2: Analyze the numerators
The numerators form an arithmetic progression (AP):
- First term \( a = 1 \)
- Common difference \( d = 3 \)

The \( n \)-th term of the AP can be expressed as:
\[ a_n = 1 + (n-1) \cdot 3 = 3n - 2 \]

### Step 3: Write the series in summation form
The series can be rewritten using the \( n \)-th term:
\[ S = \sum_{n=1}^{\infty} \frac{3n - 2}{5^{n-1}} \]

### Step 4: Split the series
We can split the series into two separate sums:
\[ S = \sum_{n=1}^{\infty} \frac{3n}{5^{n-1}} - \sum_{n=1}^{\infty} \frac{2}{5^{n-1}} \]

### Step 5: Calculate the second sum
The second sum is a geometric series:
\[ \sum_{n=1}^{\infty} \frac{2}{5^{n-1}} = 2 \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n = 2 \cdot \frac{1}{1 - \frac{1}{5}} = 2 \cdot \frac{5}{4} = \frac{10}{4} = \frac{5}{2} \]

### Step 6: Calculate the first sum
For the first sum, we use the formula for the sum of \( n \) terms multiplied by \( n \):
\[ \sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2} \]
Here, \( x = \frac{1}{5} \):
\[ \sum_{n=1}^{\infty} n \left(\frac{1}{5}\right)^{n-1} = \frac{1}{\left(1 - \frac{1}{5}\right)^2} = \frac{1}{\left(\frac{4}{5}\right)^2} = \frac{25}{16} \]
Thus:
\[ \sum_{n=1}^{\infty} \frac{3n}{5^{n-1}} = 3 \cdot \frac{25}{16} = \frac{75}{16} \]

### Step 7: Combine the results
Now substituting back into the equation for \( S \):
\[ S = \frac{75}{16} - \frac{5}{2} = \frac{75}{16} - \frac{40}{16} = \frac{35}{16} \]

### Final Answer
Thus, the sum to infinity of the series is:
\[ \boxed{\frac{35}{16}} \]