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 2 years ago
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
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.

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0ok 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

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

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

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

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

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

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0another 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

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

frx
 2 years ago
Best ResponseYou've already chosen the best response.0I 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

frx
 2 years ago
Best 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 ?\[a _{n}=\frac{ 1 }{ n }\]
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