1975 Canadian MO Problems/Problem 2

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

Page

Toolbox

Search

1975 Canadian MO Problems/Problem 2

Problem 2

Solution