how do i solve this infinite geometric series?
64/25-16/5+4-5

Answers

Answer 1

Answer: $(64)/(25)-(16)/(5)+4-5+....\n\na_1=(64)/(25);\ a_2=-(16)/(5)\n\nq=a_2:a_1\n\nq=-(16)/(5):(64)/(25)=-(16)/(5)\cdot(25)/(64)=-(5)/(4)\n\nS_n=(a_1(1-q^n))/(1-q)\n\n\nS_n=((64)/(25)(1-(-(5)/(4))^n))/(1-(-(5)/(4)))=(64)/(25)(1-(-(5)/(4))^n):(1+(5)/(4))=(64)/(25)(1-(-(5)/(4))^n):(9)/(4)\n\n=(64)/(25)(1-(-(5)/(4))^n)\cdot(4)/(9)=(256)/(225)\cdot\left(1-\left(-(5)/(4)\right)^n\right)$

Answer 2

Answer: $x= (64)/(25) - (16)/(5) +4-5+...\n \na_1= (64)/(25)\ \ \ \wedge \ \ \ a_2= - (16)/(5)\ \ \ \Rightarrow\ \ \ q= (a_2)/(a_1) = (-16)/(5):(64)/(25)= (-16)/(5)\cdot (25)/(64)=- (5)/(4) \n \nx=sum\ of\ the\ infinite\ geometric\ series\n \n$

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

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

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Help pls i need this luv u guys <33

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So this is so easy
the answer is33
pay attention what i’m saying:
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