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frx
Group Title
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.
 2 years ago
 2 years ago
frx Group Title
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.
 2 years ago
 2 years ago

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suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
we will get a0 as 0 if we put n =1
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
ok this is what we do when we have just \(a_{n+1}\) and \(a_n\) in recursion formula \[a _{n+1}a _{n}=\frac{ 1 }{ n(n+1) }\]\[a _{n}a _{n1}=\frac{ 1 }{ n(n1) }\]\[...\]\[a_2a_1=\frac{1}{2}\]and add them all
 2 years ago

suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
sorry cant find a0
 2 years ago

suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
what mukushala has done is correct then you will get a(n+1) = 1  sum of terms of rhs
 2 years ago

suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
for finding the sum of terms on rhs write each term like this\[\frac{ 1 }{(n)(n1) } = \frac{ n  (n1) }{(n)(n1) } = \frac{ 1 }{ n1 }  \frac{ 1 }{ n }\]
 2 years ago

suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
now when you add consecutive terms will get cancelled and you will be left with \[1  \frac{ 1 }{ n+1 }\]
 2 years ago

suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
did you understand frx?
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Well, not really yet but thank you, will have to sit down and think about what you guys just wrote:)
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
I can't see what the closed formula is? Is it \[1\frac{ 1 }{ n+1}\]?
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
yeah what mukushla is saying is better i think
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 1 }{ n+1 }\frac{ 1 }{ n }\] Looking at this expression, why should i guess the closed form is ?\[a _{n}=\frac{ 1 }{ n }\]
 2 years ago
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