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

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

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

$$\sum _{\mathrm{n}=2}^{24}\frac{250}{{(1+\mathrm{x})}^{\mathrm{n}}}=5000\Rightarrow \sum _{\mathrm{n}=3}^{25}\frac{1}{{\mathrm{x}}^{\mathrm{n}-1}}=\frac{5000}{250}=\frac{500}{25}=20\Rightarrow$$$$\sum _{\mathrm{n}=3}^{25}{\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{n}-1}=20\Rightarrow \sum _{\mathrm{n}=1}^{25}{\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{n}-1}-1-\frac{1}{\mathrm{x}}=20\Rightarrow$$$$\sum _{\mathrm{n}=0}^{24}{\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{n}}-1-\frac{1}{\mathrm{x}}=20\Rightarrow \frac{1-{\left(\frac{1}{\mathrm{x}}\right)}^{25}}{1-\frac{1}{\mathrm{x}}}-1-\frac{1}{\mathrm{x}}=20\Rightarrow$$$$\frac{\mathrm{x}(1-\frac{1}{{\mathrm{x}}^{25}})}{\mathrm{x}-1}-(1+\frac{1}{\mathrm{x}})=20\Rightarrow$$$$\frac{\mathrm{x}-\frac{1}{{\mathrm{x}}^{24}}-(\mathrm{x}-1)(1+\frac{1}{\mathrm{x}})}{\mathrm{x}-1}=20\Rightarrow$$$$\mathrm{x}-\frac{1}{{\mathrm{x}}^{24}}-(\mathrm{x}+1-1-\frac{1}{\mathrm{x}})=20(\mathrm{x}-1)\Rightarrow$$$$\mathrm{x}-\frac{1}{{\mathrm{x}}^{24}}-\mathrm{x}+\frac{1}{\mathrm{x}}=20\mathrm{x}-20\Rightarrow$$$$\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^{24}}-20\mathrm{x}+20=0\Rightarrow$$$${\mathrm{x}}^{23}-1-{20\mathrm{x}}^{25}+{20\mathrm{x}}^{24}=0\Rightarrow$$$${20\mathrm{x}}^{25}-{20\mathrm{x}}^{24}-{\mathrm{x}}^{23}+1=0\mathrm{rest}\mathrm{to}\mathrm{solve}\mathrm{this}\mathrm{equation}...$$

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