2019-07-03T05:16:40-04:00
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
2019-07-05T10:41:33-0400
1 / ( 1 − x ) = 1 + x + x 2 + . . . + x n = ∑ n = 0 ∞ x n , ∣ x ∣ < 1 1/(1-x)=1+x+x^2+...+x^n=\sum_{n=0} ^\infin x^n, |x|<1 1/ ( 1 − x ) = 1 + x + x 2 + ... + x n = n = 0 ∑ ∞ x n , ∣ x ∣ < 1 Then
1 / ( 1 + x ) = 1 / ( 1 − ( − x ) ) = 1/(1+x)=1/(1-(-x))= 1/ ( 1 + x ) = 1/ ( 1 − ( − x )) =
= 1 − x + x 2 + . . . + ( − 1 ) n x n = ∑ n = 0 ∞ ( − 1 ) n x n , ∣ x ∣ < 1 =1-x+x^2+...+(-1)^nx^n=\sum_{n=0} ^\infin (-1)^nx^n, |x|<1 = 1 − x + x 2 + ... + ( − 1 ) n x n = n = 0 ∑ ∞ ( − 1 ) n x n , ∣ x ∣ < 1 If
x = 3 r x=3r x = 3 r then
∑ n = 0 ∞ ( − 1 ) n 3 n r n = 1 / ( 1 + 3 r ) \sum_{n=0}^\infin (-1)^n 3^nr^n=1/(1+3r) n = 0 ∑ ∞ ( − 1 ) n 3 n r n = 1/ ( 1 + 3 r ) for
∣ 3 r ∣ < 1 ⇒ − 1 / 3 < r < 1 / 3. |3r|<1 \Rightarrow - 1/3<r<1/3. ∣3 r ∣ < 1 ⇒ − 1/3 < r < 1/3.
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