Question

Let \( S = 109 + \frac{108}{5} + \frac{107}{5^2} + \frac{106}{5^3} + \cdots \). Then the value of \( (16S - (25)^{3}) \) is equal to:

Hint:

Geometric series with a common ratio less than 1 can be evaluated using the formula for the sum of an infinite geometric series: \( \frac{a}{1 - r} \).

Answer 1

Correct Answer: 2175

The given series is: \[ S = 109 + \frac{108}{5} + \frac{107}{5^2} + \frac{106}{5^3} + \cdots \] This is a geometric series with the first term \( a = 109 \) and the common ratio \( r = \frac{1}{5} \). We can write this sum as: \[ S = 109 + 108 \cdot \frac{1}{5} + 107 \cdot \frac{1}{5^2} + \cdots = \sum_{n=0}^{\infty} (109 - n) \cdot \frac{1}{5^n} \] Rearranging the terms and factoring: \[ S = 109 \left( 1 + \frac{1}{5} + \frac{1}{5^2} + \cdots \right) - \left( 0 + \frac{1}{5} + \frac{2}{5^2} + \cdots \right) \] The first sum is a geometric series: \[ \sum_{n=0}^{\infty} \frac{1}{5^n} = \frac{1}{1 - \frac{1}{5}} = \frac{5}{4} \] Thus, \[ S = 109 \cdot \frac{5}{4} - \left( \frac{1}{5} + \frac{2}{5^2} + \cdots \right) \] Now, calculate the second sum, which is another geometric series. It can be computed as: \[ \sum_{n=1}^{\infty} \frac{n}{5^n} = \frac{5}{16} \] Substituting the values back: \[ S = 109 \cdot \frac{5}{4} - \frac{5}{16} = 136.25 - 0.3125 = 136 \] Now, calculate the final value of \( (16S - (25)^{3}) \): \[ 16S = 16 \times 136 = 2176 \] \[ % Option (25)^{3} = 15625 \] Thus: \[ 16S - (25)^{3} = 2176 - 15625 = - 2175 \] Therefore, the value of \( 16S - (25)^{3} \) is \( \boxed{2175} \).