Verified Answer :

\(\frac{64}{25}-\frac{16}{5}+4-5+....\\\\a_1=\frac{64}{25};\ a_2=-\frac{16}{5}\\\\q=a_2:a_1\\\\q=-\frac{16}{5}:\frac{64}{25}=-\frac{16}{5}\cdot\frac{25}{64}=-\frac{5}{4}\\\\S_n=\frac{a_1(1-q^n)}{1-q}\\\\\\S_n=\frac{\frac{64}{25}(1-(-\frac{5}{4})^n)}{1-(-\frac{5}{4})}=\frac{64}{25}(1-(-\frac{5}{4})^n):(1+\frac{5}{4})=\frac{64}{25}(1-(-\frac{5}{4})^n):\frac{9}{4}\\\\=\frac{64}{25}(1-(-\frac{5}{4})^n)\cdot\frac{4}{9}=\frac{256}{225}\cdot\left(1-\left(-\frac{5}{4}\right)^n\right)\)

kate200468

\(x= \frac{64}{25} - \frac{16}{5} +4-5+...\\ \\a_1= \frac{64}{25}\ \ \ \wedge \ \ \ a_2= - \frac{16}{5}\ \ \ \Rightarrow\ \ \ q= \frac{a_2}{a_1} = \frac{-16}{5}:\frac{64}{25}= \frac{-16}{5}\cdot \frac{25}{64}=- \frac{5}{4} \\ \\x=sum\ of\ the\ infinite\ geometric\ series\\ \\\)

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

\(n=2k+1\ \ \ \Rightarrow\ \ \ \lim_{n \to \infty} (- \frac{5}{4} )^n=-\infty\ \ \Rightarrow\ \ \ x \rightarrow -\infty\)

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