Question #91365
Q. Choose the correct answer.
Q. Which of the following is the sum of infinite series ∑_(n=0)^∞▒〖(-1)〗^n 3^n r^n?
a. 1/(1+3r) for -1/3<r<1/3
b. . 1/(1-3r) for -1/3<r<1/3
. 3/(1-3r) for -1/3<r<1/3
d. ∞
1
Expert's answer
2019-07-05T10:41:33-0400
1/(1x)=1+x+x2+...+xn=n=0xn,x<11/(1-x)=1+x+x^2+...+x^n=\sum_{n=0} ^\infin x^n, |x|<1

- the geometric series.

Then


1/(1+x)=1/(1(x))=1/(1+x)=1/(1-(-x))=

=1x+x2+...+(1)nxn=n=0(1)nxn,x<1=1-x+x^2+...+(-1)^nx^n=\sum_{n=0} ^\infin (-1)^nx^n, |x|<1

If

x=3rx=3r

then


n=0(1)n3nrn=1/(1+3r)\sum_{n=0}^\infin (-1)^n 3^nr^n=1/(1+3r)

for

3r<11/3<r<1/3.|3r|<1 \Rightarrow - 1/3<r<1/3.


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