If `S=7/5+9/5^2+13/5^3+ . . . .` then find value of 160S

Answer

To solve the problem \( S = \frac{7}{5} + \frac{9}{5^2} + \frac{13}{5^3} + \ldots \), we will first express \( S \) in a more manageable form and then find its value step by step.

### Step 1: Identify the series
The series can be rewritten as:
\[ S = \sum_{n=1}^{\infty} \frac{a_n}{5^n} \]
where \( a_n \) is the sequence of numerators. Observing the numerators:
- For \( n=1 \), \( a_1 = 7 \)
- For \( n=2 \), \( a_2 = 9 \)
- For \( n=3 \), \( a_3 = 13 \)

The pattern shows that \( a_n = 4 + 2n \).

### Step 2: Rewrite \( S \)
Now we can express \( S \) as:
\[ S = \sum_{n=1}^{\infty} \frac{4 + 2n}{5^n} = \sum_{n=1}^{\infty} \frac{4}{5^n} + \sum_{n=1}^{\infty} \frac{2n}{5^n} \]

### Step 3: Calculate the first sum
The first sum is a geometric series:
\[ \sum_{n=1}^{\infty} \frac{4}{5^n} = 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = 4 \cdot \frac{\frac{1}{5}}{1 - \frac{1}{5}} = 4 \cdot \frac{\frac{1}{5}}{\frac{4}{5}} = 4 \cdot \frac{1}{4} = 1 \]

### Step 4: Calculate the second sum
For the second sum, we use the formula for the sum of \( n x^n \):
\[ \sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2} \]
Setting \( x = \frac{1}{5} \):
\[ \sum_{n=1}^{\infty} n \left(\frac{1}{5}\right)^n = \frac{\frac{1}{5}}{\left(1 - \frac{1}{5}\right)^2} = \frac{\frac{1}{5}}{\left(\frac{4}{5}\right)^2} = \frac{\frac{1}{5}}{\frac{16}{25}} = \frac{25}{80} = \frac{5}{16} \]
Thus,
\[ \sum_{n=1}^{\infty} \frac{2n}{5^n} = 2 \cdot \frac{5}{16} = \frac{10}{16} = \frac{5}{8} \]

### Step 5: Combine the results
Now we can combine both sums:
\[ S = 1 + \frac{5}{8} = \frac{8}{8} + \frac{5}{8} = \frac{13}{8} \]

### Step 6: Find \( 160S \)
Now we need to find \( 160S \):
\[ 160S = 160 \cdot \frac{13}{8} = 20 \cdot 13 = 260 \]

### Final Answer
Thus, the value of \( 160S \) is:
\[ \boxed{260} \]