Find-x-if-n-2-24-250-1-x-n-5000-

Question Number 141715 by SOMEDAVONG last updated on 22/May/21

Find x if \underset{n = 2}{\sum^{24}} \frac{250}{{(1 + x)}^{n}} = 5000 .

Answered by mathmax by abdo last updated on 22/May/21

\sum_{n = 2}^{24} \frac{250}{{(1 + x)}^{n}} = 5000 \Rightarrow \sum_{n = 3}^{25} \frac{1}{x^{n - 1}} = \frac{5000}{250} = \frac{500}{25} = 20 \Rightarrow

\sum_{n = 3}^{25} {(\frac{1}{x})}^{n - 1} = 20 \Rightarrow \sum_{n = 1}^{25} {(\frac{1}{x})}^{n - 1} - 1 - \frac{1}{x} = 20 \Rightarrow

\sum_{n = 0}^{24} {(\frac{1}{x})}^{n} - 1 - \frac{1}{x} = 20 \Rightarrow \frac{1 - {(\frac{1}{x})}^{25}}{1 - \frac{1}{x}} - 1 - \frac{1}{x} = 20 \Rightarrow

\frac{x (1 - \frac{1}{x^{25}})}{x - 1} - (1 + \frac{1}{x}) = 20 \Rightarrow

\frac{x - \frac{1}{x^{24}} - (x - 1) (1 + \frac{1}{x})}{x - 1} = 20 \Rightarrow

x - \frac{1}{x^{24}} - (x + 1 - 1 - \frac{1}{x}) = 20 (x - 1) \Rightarrow

x - \frac{1}{x^{24}} - x + \frac{1}{x} = 20 x - 20 \Rightarrow

\frac{1}{x} - \frac{1}{x^{24}} - 20 x + 20 = 0 \Rightarrow

x^{23} - 1 - {20 x}^{25} + {20 x}^{24} = 0 \Rightarrow

{20 x}^{25} - {20 x}^{24} - x^{23} + 1 = 0 rest to solve this equation \dots