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October 2, 2014

October 2, 2014

Posted by **nikki** on Thursday, January 20, 2011 at 3:17pm.

Y1 = 7 over Σ 15/4(pi/10)^n

Y2= 11 over Σ 17/4(4/13)^n

- MATH -
**Reiny**, Monday, January 24, 2011 at 10:38amdid you mean

Σ ((15/4) (π/10)^n , where n goes from 1 to 7 ?

of so, then you would have

(15/4) [ π/10 + π^2/100 + ... + π^7/10^7)

the [..] is a geometric series where a=π/10 and r = π/10 for 7 terms

the value of the [..]

= (π/10)((π/10)^7 - 1)/(π/10 - 1) = appr. .4579

so

Σ ((15/4) (π/10)^n , where n goes from 1 to 7

= (15/4)(.4579) = 1.717

evaluate the second part the same way, and compare sums

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