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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.
 one year ago
 one year 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.
 one year ago
 one year 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
 one year 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
 one year ago

suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
sorry cant find a0
 one year 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
 one year 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\[\frac{ 1 }{(n)(n1) } = \frac{ n  (n1) }{(n)(n1) } = \frac{ 1 }{ n1 }  \frac{ 1 }{ n }\]
 one year 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 \[1  \frac{ 1 }{ n+1 }\]
 one year ago

suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
did you understand frx?
 one year 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:)
 one year ago

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

suvesh253 Group TitleBest ResponseYou've already chosen the best response.0
yeah what mukushla is saying is better i think
 one year 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
 one year 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 ?\[a _{n}=\frac{ 1 }{ n }\]
 one year ago
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