1973 Canadian MO Problems/Problem 7

Problem

$\text{(i)}$ Observe that $\frac{1}{1}=$ $\frac{1}{2}+$ $\frac{1}{2};$ $\quad$ $\frac{1}{2}=$ $\frac{1}{3}+$ $\frac{1}{6};$ $\quad \frac{1}{3}=$ $\frac{1}{4}+\frac{1}{12};\quad...$ State a general law suggested by these examples, and prove it.

$\text{(ii)}$ Prove that for any integer $n$ greater than $1$ there exist positive integers $i$ and $j$ such that $\frac{1}{n}= \frac{1}{i(i+1)}+\frac{1}{(i+1)(i+2)}+\frac{1}{(i+2)(i+3)}+\cdots+\frac{1}{j(j+1)}.$

Solution 1 (Easiest)

$\text{i}$ : We see that $\frac{1}{n(n+1)}+\frac{1}{n+1} = \frac{1}{n}$ . By simply evaluating the left hand side using common denominators, we see that both sides are equal. So therefore we have proved the law.

$\text{ii}$ : Using partial fraction decomposition, let's change the original expression to:

$\frac{1}{i}-\frac{1}{i+1}+\frac{1}{i+1}-\frac{1}{i+2}+...+\frac{1}{j}-\frac{1}{j+1}=\frac{1}{i}-\frac{j+1}=\frac{1}{n}$ .

Therefore, $\frac{1}{i}=\frac{1}{n}+\frac{1}{j+1}$ .

Let $i=(j+1)(j+2), n=j+2$ , then we have satisfied the previous identity in part $\text{i}$ . Therefore, we have solved the problem.

~hastapasta

Solution 2

$\text{(i)}$ We see that:

$\frac{1}{n(n+1)}+\frac{1}{n+1} = \frac{1}{n}$

We prove this by induction. Let $P(n)=\frac{1}{n(n+1)}+\frac{1}{n+1}= \frac{1}{n}.$ Base case: $n = 1 \Rightarrow P(1):\text{LHS} = 1, \text{while RHS} = \frac{1}{1} = 1.$ Therefore, $P(1)$ is true. Now, assume that $P(n)$ is true for some $n=k$ . Then:

$\begin{matrix} \text{LHS of } P(k+1) &=& \frac{1}{(k+1)((k+1)+1)}+\frac{1}{(k+1)+1}\\ &=& \frac{1}{(k+1)(k+2)}+\frac{1}{k+2}\\ &=& \frac{1}{(k+1)(k+2)}+\frac{k+1}{(k+2)(k+1)}\\ &=& \frac{k+2}{(k+1)(k+2)}\\ &=& \frac{1}{k+1} &=& \text{RHS of } P(k+1)\\\end{matrix}$

Thus, by induction, the formula holds for all $n \in \mathbb{Z^{+}}$

$\text{(ii)}$ Incomplete

1973 Canadian MO (Problems)
Preceded by Problem 6	1 • 2 • 3 • 4 • 5	Followed by Problem 1

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1973 Canadian MO Problems/Problem 7

Contents

Problem

Solution 1 (Easiest)

Solution 2

See also