anonymous
  • anonymous
How do i find the sum?? http://i.imgur.com/tBwQ0Fi.png I know to represent it as a geometric series but I am unsure what to do with the ^n-1
Mathematics
chestercat
  • chestercat
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Astrophysics
  • Astrophysics
We can use geometric series test, so first lets find the partial sum for \[\sum_{n=1}^{\infty} \frac{ (-4)^{n-1} }{ 5^n }\] so we let n = 1
Astrophysics
  • Astrophysics
\[a_1 = \frac{ (-4)^{1-1} }{ 5^1 } = \frac{ (-4)^0 }{ 5 } = \frac{ 1 }{ 5 }\] and we know if |r|<1 it is then convergent to find r we can use \[a_1r=a_2 \implies \frac{ 1 }{ 5 }r=-\frac{ 4 }{ 25 } \implies r = -\frac{ 20 }{ 25 } = -\frac{ 4 }{ 5 }\] \[\frac{ a_1 }{ 1-r } = \frac{ 1/5 }{ 1+4/5 } = \frac{ 1 }{ 9 }\] therefore it's convergent
Astrophysics
  • Astrophysics
Your sum is 1/9

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anonymous
  • anonymous
Oh i see we manipulated a2/a1=r
Astrophysics
  • Astrophysics
Yup
anonymous
  • anonymous
Why did we use n=1?
Astrophysics
  • Astrophysics
That gives us our first term a_1
Astrophysics
  • Astrophysics
Remember r is the common ratio here
anonymous
  • anonymous
Oh right n=1 in the sum so its our first term, i keep on thinking it starts at zero. Thanks
Astrophysics
  • Astrophysics
Np :)

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