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# math problem?

((1/2) * 1) - ((3/10) * 2) + ((9/50) * 3) - ((27/250) * 4) + ...

### 2 Answers

- PinkgreenLv 71 month ago
Let the sum of the series be S such that

......inf.

S=SUM{[(-1)^(n+1)]n(0.6)^(n-1)}/2

.....n=1

Consider

.......inf.

S=SUM{[(-1)^(n+1)]nx^(n-1)}/2

.....n=1

=>

S=(d/dx){x-x^2+x^3-......}/2

=>

S=(1/2)(d/dx){x/(1+x)}, for |x|<1

=>

S=1/[2(1+x)^2]

Let x=0.6, then

S=1/[2(1.6)^2]

=>

S=25/128

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- 1 month ago
3^(n - 1) * n / (2 * 5^(n - 1)) =>

(3/5)^(n - 1) * (1/2) * n

Let's worry about the 1/2 bit later

S = (3/5)^0 * 1 + (3/5)^1 * 2 + (3/5)^2 * 3 + ....

S * (3/5) = (3/5)^1 * 1 + (3/5)^2 * 2 + (3/5)^3 * 3 + ....

S - S * (3/5) = (3/5)^0 * 1 + (3/5)^1 * 2 - (3/5)^1 * 1 + (3/5)^2 * 3 - (3/5)^2 * 2 + ....

S * (1 - 3/5) = 1 + (3/5) * (2 - 1) + (3/5)^2 * (3 - 2) + (3/5)^3 * (4 - 3) + ...

S * (2/5) = 1 + (3/5) * 1 + (3/5)^2 * 1 + (3/5)^3 * 1 + ....

T = (3/5) + (3/5)^2 + (3/5)^3 + ...

T = (3/5) + (3/5) * ((3/5) + (3/5)^2 + (3/5)^3 + ...)

T = (3/5) + (3/5) * T

T - (3/5) * T = (3/5)

T * (1 - 3/5) = (3/5)

T * (2/5) = (3/5)

2T = 3

T = 3/2

S * (2/5) = 1 + T

S * (2/5) = 1 + 3/2

S * (2/5) = 5/2

S = (5/2) * (5/2)

S = 25/4

Multiply by 1/2 from the beginning

(1/2) * (25/4) =>

25/8 =>

3.125

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