Find the sum of the series `1+4/5+7/(5^2)+10/(5^3)+...`

Options

A. `7//16`
B. `5//16`
C. `105//64`
D. `35//16`

Explanation

To find the sum 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
The series can be expressed as:
\[ S = 1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \ldots \]
We notice that the numerators form an arithmetic sequence: \( 1, 4, 7, 10, \ldots \) where the \( n \)-th term can be expressed as \( 3n - 2 \) for \( n = 1, 2, 3, \ldots \).

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

### Step 3: Calculate the first sum
The first sum can be calculated using the formula for the sum of an infinite series:
\[ \sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2} \quad \text{for } |x| < 1 \]
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 4: 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 5: Combine the results
Now, substituting back into our expression for \( S \):
\[ S = \frac{75}{16} - \frac{5}{2} \]
To combine these, we convert \( \frac{5}{2} \) to have a common denominator of 16:
\[ \frac{5}{2} = \frac{5 \cdot 8}{2 \cdot 8} = \frac{40}{16} \]
Thus,
\[ S = \frac{75}{16} - \frac{40}{16} = \frac{35}{16} \]

### Final Answer
The sum of the series is:
\[ \boxed{\frac{35}{16}} \]