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Difference between revisions of "1975 Canadian MO Problems/Problem 2"
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== Solution ==  
 
== Solution ==  
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I claim <math>a_n=\frac{1}{(n)(n+1)}</math> with a proof by induction. First we can use partial fraction decomposition to rewrite <math>\frac{1}{(n)(n+1)}</math> as <math>\frac{A}{n}+\frac{B}{n+1}</math>. We have <cmath>\frac{An+A+Bn}{(n)(n+1)}=\frac{1}{(n)(n+1)}.</cmath> We can set coefficients equal, <math>A+B=0</math> and <math>A=1</math>. Now, <cmath>\frac{1}{(n)(n+1)}=\frac{1}{n}-\frac{1}{n+1}.</cmath>
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'''Base Case:''' If <math>n=1</math>, then <math>\frac{1}{(1)(1+1)}=\frac{1}{2}.</math> So, <math>a_n=\frac{1}{(n)(n+1)}</math> when <math>n=1</math>.
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'''Inductive Step:''' Suppose conclusion is true for <math>n=k</math>, such that we have <math>a_k=\frac{1}{k}-\frac{1}{k+1}.</math> We also have <math>a_1+a_2+\cdots+a_k=k^2a_k.</math> Add <math>a_{k+1}</math> to both sides. The left side becomes <math>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{k}-\frac{1}{k+1}+a_{k+1}</math> which is a telescoping series equal to <math>1-\frac{1}{k+1}+a_{k+1}=\frac{k}{k+1}+a_{k+1}</math>. Now, we have <cmath>a_1+a_2+\cdots+a_k+a_{k+1}=\frac{k}{k+1}+a_{k+1}=(k+1)^2a_{k+1}.</cmath> We have <cmath>\frac{k}{k+1}+a_{k+1}=(k+1)^2a_{k+1}\implies \frac{k}{k+1}=(k^2+2k+1-1)(a_{k+1}) \implies \left(\frac{k}{k+1}\right)\left(\frac{1}{(k)(k+2)}\right)=a_{k+1} \implies a_{k+1}=\frac{1}{(k+1)(k+2)},</cmath> thus the conclusion being true for <math>n=k</math>, implies that it holds for <math>n=k+1</math>, so our induction is complete.<math>\blacksquare</math>
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{{Old CanadaMO box|num-b=1|num-a=3|year=1975}}
 
{{Old CanadaMO box|num-b=1|num-a=3|year=1975}}

Latest revision as of 17:42, 20 June 2018

Problem 2

A sequence of numbers $a_1, a_2, a_3, \dots$ satisfies

(i) $a_1 = \frac{1}{2}$
(ii) $a_1+a_2+\cdots+a_n=n^2a_n\quad(n\ge1).$

Determine the value of $a_n\quad(n\ge1).$

Solution

I claim $a_n=\frac{1}{(n)(n+1)}$ with a proof by induction. First we can use partial fraction decomposition to rewrite $\frac{1}{(n)(n+1)}$ as $\frac{A}{n}+\frac{B}{n+1}$. We have \[\frac{An+A+Bn}{(n)(n+1)}=\frac{1}{(n)(n+1)}.\] We can set coefficients equal, $A+B=0$ and $A=1$. Now, \[\frac{1}{(n)(n+1)}=\frac{1}{n}-\frac{1}{n+1}.\]


Base Case: If $n=1$, then $\frac{1}{(1)(1+1)}=\frac{1}{2}.$ So, $a_n=\frac{1}{(n)(n+1)}$ when $n=1$.


Inductive Step: Suppose conclusion is true for $n=k$, such that we have $a_k=\frac{1}{k}-\frac{1}{k+1}.$ We also have $a_1+a_2+\cdots+a_k=k^2a_k.$ Add $a_{k+1}$ to both sides. The left side becomes $\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{k}-\frac{1}{k+1}+a_{k+1}$ which is a telescoping series equal to $1-\frac{1}{k+1}+a_{k+1}=\frac{k}{k+1}+a_{k+1}$. Now, we have \[a_1+a_2+\cdots+a_k+a_{k+1}=\frac{k}{k+1}+a_{k+1}=(k+1)^2a_{k+1}.\] We have \[\frac{k}{k+1}+a_{k+1}=(k+1)^2a_{k+1}\implies \frac{k}{k+1}=(k^2+2k+1-1)(a_{k+1}) \implies \left(\frac{k}{k+1}\right)\left(\frac{1}{(k)(k+2)}\right)=a_{k+1} \implies a_{k+1}=\frac{1}{(k+1)(k+2)},\] thus the conclusion being true for $n=k$, implies that it holds for $n=k+1$, so our induction is complete.$\blacksquare$


1975 Canadian MO (Problems)
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 3


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