To solve the problem, note that \( S = 109 + \frac{108}{5} + \frac{107}{5^2} + \frac{106}{5^3} + \cdots \) is an infinite geometric series where the first term \( a = 109 \) and the common ratio \( r = \frac{1}{5} \). The general term can be expressed as \( a_n = (110-n)\cdot \left(\frac{1}{5}\right)^{n-1} \).
The sum \( S \) can be rewritten as a series:
\( S = \sum_{n=1}^{\infty} (110-n)\cdot \left(\frac{1}{5}\right)^{n-1} \).
This series is the sum of two geometric series:
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 \( \boxed{2175} \).
If \( A = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} \), then \( A^{50} \) is:
The range of the function \( f(x) = \sin^{-1}(x - \sqrt{x}) \) is equal to?
The function \( f(x) = \tan^{-1} (\sin x + \cos x) \) is an increasing function in: