Answered

Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

how do i solve this infinite geometric series?

64/25-16/5+4-5

Sagot :

[tex]\frac{64}{25}-\frac{16}{5}+4-5+....\\\\a_1=\frac{64}{25};\ a_2=-\frac{16}{5}\\\\q=a_2:a_1\\\\q=-\frac{16}{5}:\frac{64}{25}=-\frac{16}{5}\cdot\frac{25}{64}=-\frac{5}{4}\\\\S_n=\frac{a_1(1-q^n)}{1-q}\\\\\\S_n=\frac{\frac{64}{25}(1-(-\frac{5}{4})^n)}{1-(-\frac{5}{4})}=\frac{64}{25}(1-(-\frac{5}{4})^n):(1+\frac{5}{4})=\frac{64}{25}(1-(-\frac{5}{4})^n):\frac{9}{4}\\\\=\frac{64}{25}(1-(-\frac{5}{4})^n)\cdot\frac{4}{9}=\frac{256}{225}\cdot\left(1-\left(-\frac{5}{4}\right)^n\right)[/tex]
kate200468
[tex]x= \frac{64}{25} - \frac{16}{5} +4-5+...\\ \\a_1= \frac{64}{25}\ \ \ \wedge \ \ \ a_2= - \frac{16}{5}\ \ \ \Rightarrow\ \ \ q= \frac{a_2}{a_1} = \frac{-16}{5}:\frac{64}{25}= \frac{-16}{5}\cdot \frac{25}{64}=- \frac{5}{4} \\ \\x=sum\ of\ the\ infinite\ geometric\ series\\ \\[/tex]

[tex]x= \lim_{n \to \infty} a_1\cdot \frac{1-q^n}{1-q} = \lim_{n \to \infty} \frac{64}{25} \cdot \frac{1-(- \frac{5}{4})^n }{1+ \frac{5}{4} } =\\ \\= \lim_{n \to \infty} \frac{64}{25} \cdot \frac{4}{9} \cdot [1-(- \frac{5}{4})^n]= \frac{256}{225} +\frac{256}{225}\cdot \lim_{n \to \infty} (- \frac{5}{4} )^n\\ \\n=2k\ \ \Rightarrow\ \ \ \lim_{n \to \infty} (- \frac{5}{4} )^n=+\infty\ \ \Rightarrow\ \ \ x\rightarrow+\infty\\ \\[/tex]

[tex]n=2k+1\ \ \ \Rightarrow\ \ \ \lim_{n \to \infty} (- \frac{5}{4} )^n=-\infty\ \ \Rightarrow\ \ \ x \rightarrow -\infty[/tex]
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.