# does this make sense to you?

just wondering why muliplying this is true.

(0.0001) ∑ k=0 to infinity [(.99)^2]^k = 0.0001 / (1-(.99)^2)

so its the .0001 multiplied by the sum from 0 to infinity multiplied by [(.99)^2]^k

why does the left side equal the right side, i dont get it

### 4 Answers

- KimLv 61 decade agoFavorite Answer
(0.0001) ∑ k=0 to infinity [(.99)^2]^k = 0.0001 / (1-(.99)^2)

an infinite geometric series is S∞ = a1 / (1-r ) where r is between 0 and 1

What you have here looks like a geometric series. See

http://www.mathwords.com/i/infinite_geometric_seri...

a1 = 0.0001 and r = (.99)^2

- hippoLv 61 decade ago
(0.0001) â k=0 to infinity [(.99)^2]^k = 0.0001 / (1-(.99)^2)

â k=0 to infinity [(.99)^2]^k = 1 / (1-(.99)^2)

â k=0 to infinity [(.99)^(2k)] = 1 / (1-(.99)^2)

(.99)^0 + (.99)^2 + (.99)^4 ... = 1 / (1-(.99)^2)

(1 - (.99)^2)[(.99)^0 + (.99)^2 + (.99)^4 ...] = 1

[(.99)^0 + (.99)^2 + (.99)^4 ...] - (.99)^2[(.99)^0 + (.99)^2 + (.99)^4 ...] = 1

[(.99)^0 + (.99)^2 + (.99)^4 +...] - [(.99)^2 + (.99)^4 + ...] = 1

[(.99)^0] = 1

1 = 1

TRUE