anonymous
  • anonymous
The number sequence \[a _{0},a _{1},a2...\] is defined by the recursion formula: \[a _{1}=1\] \[a _{n+1}=a _{n}−\frac{ 1 }{ n(n+1) },n≥1\] Find the closed formula for the sequence and prove it by induction.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
we will get a0 as 0 if we put n =1
anonymous
  • anonymous
ok this is what we do when we have just \(a_{n+1}\) and \(a_n\) in recursion formula \[a _{n+1}-a _{n}=\frac{ 1 }{ n(n+1) }\]\[a _{n}-a _{n-1}=\frac{ 1 }{ n(n-1) }\]\[...\]\[a_2-a_1=\frac{1}{2}\]and add them all
anonymous
  • anonymous
sorry cant find a0

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anonymous
  • anonymous
what mukushala has done is correct then you will get a(n+1) = 1 - sum of terms of rhs
anonymous
  • anonymous
for finding the sum of terms on rhs write each term like this\[\frac{ 1 }{(n)(n-1) } = \frac{ n - (n-1) }{(n)(n-1) } = \frac{ 1 }{ n-1 } - \frac{ 1 }{ n }\]
anonymous
  • anonymous
now when you add consecutive terms will get cancelled and you will be left with \[1 - \frac{ 1 }{ n+1 }\]
anonymous
  • anonymous
did you understand frx?
anonymous
  • anonymous
Well, not really yet but thank you, will have to sit down and think about what you guys just wrote:)
anonymous
  • anonymous
I can't see what the closed formula is? Is it \[1-\frac{ 1 }{ n+1}\]?
anonymous
  • anonymous
another way to seeing this is writing it in the form\[a _{n+1}-a _{n}=-\frac{1}{n(n+1)}=\frac{ 1 }{ n+1 }-\frac{1}{n}\]and guessing closed form like \[a_n=\frac{1}{n}\]then prove it by induction because it says find the closed formula(even by guessing) then prove by induction
anonymous
  • anonymous
yeah what mukushla is saying is better i think
anonymous
  • anonymous
I really don't get it :/ I get the first part when subtractin a_n from both sides but is 1/(n+1)-1/n equal to the first ecpression, I can't see it
anonymous
  • anonymous
\[\frac{ 1 }{ n+1 }-\frac{ 1 }{ n }\] Looking at this expression, why should i guess the closed form is ?\[a _{n}=\frac{ 1 }{ n }\]

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