Question

how do i solve this infinite geometric series?

64/25-16/5+4-5

Answer 1

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


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


`n=2k+1\ \ \ \Rightarrow\ \ \ \lim_(n \to \infty) (- (5)/(4) )^n=-\infty\ \ \Rightarrow\ \ \ x \rightarrow -\infty`

Answer 2

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