Question

If \( S = \frac{7}{5} + \frac{9}{5^2} + \frac{13}{5^3} + \frac{19}{5^4} + ... \), then 160 S is equal to _________.

Hint:

For AGP sums, if the numerator differences are not constant, subtract again! Usually, second-order AGPs reduce to standard ones after the first subtraction.

Answer 1

Correct Answer: 305

Step 1: Understanding the Concept:
This series is an Arithmetico-Geometric Progression (AGP). The numerators follow a pattern of differences that eventually becomes an Arithmetic Progression. We use the standard method of subtracting a shifted, multiplied version of the sum to simplify it.
Step 2: Detailed Explanation:
Let \( S = \frac{7}{5} + \frac{9}{5^2} + \frac{13}{5^3} + \frac{19}{5^4} + \dots \) (1)
Multiply by \( \frac{1}{5} \): \( \frac{S}{5} = \frac{7}{5^2} + \frac{9}{5^3} + \frac{13}{5^4} + \dots \) (2)
Subtract (2) from (1): \[ \frac{4S}{5} = \frac{7}{5} + \frac{2}{5^2} + \frac{4}{5^3} + \frac{6}{5^4} + \dots \] \[ \frac{4S}{5} = \frac{7}{5} + \frac{2}{25} \left( 1 + \frac{2}{5} + \frac{3}{5^2} + \dots \right) \]
The term in parentheses is a standard AGP sum: \( \sum_{n=1}^\infty n \cdot r^{n-1} = \frac{1}{(1-r)^2} \). Here \( r = \frac{1}{5} \), so: \[ \left( 1 + \frac{2}{5} + \frac{3}{5^2} + \dots \right) = \frac{1}{(1 - 1/5)^2} = \frac{1}{(4/5)^2} = \frac{25}{16} \]
Now, substitute back: \[ \frac{4S}{5} = \frac{7}{5} + \frac{2}{25} \cdot \frac{25}{16} = \frac{7}{5} + \frac{1}{8} = \frac{56 + 5}{40} = \frac{61}{40} \] \[ S = \frac{61}{40} \cdot \frac{5}{4} = \frac{61}{32} \]
Calculate 160 S: \[ 160 \cdot \frac{61}{32} = 5 \cdot 61 = 305 \]
Step 3: Final Answer:
The value of 160 S is 305.