**Problem**: (1969 Canada - #1) If $ \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} $ and $ p_1, p_2, p_3 $ are not all zero, show that for all $ n \in \mathbb{N} $,

$ \left(\frac{a_1}{b_1}\right)^n = \frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n} $.

**Solution**: Well, a good first step seems to be expanding it out. We want to show

$ p_1(a_1b_1)^n+p_2(a_1b_2)^n+p_3(a_1b_3)^n = p_1(a_1b_1)^n+p_2(a_2b_1)^n+p_3(a_3b_1)^n $,

which reduces to

$ p_2(a_1b_2)^n+p_3(a_1b_3)^n = p_2(a_2b_1)^n+p_3(a_3b_1)^n $.

However, noting that $ \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \Rightarrow a_1b_2 = a_2b_1 $ and $ a_1b_3 = a_3b_1 $, we simply substitute and the problem is solved. QED.

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Comment: Wow. Back in the old days, this qualified for an olympiad problem. It looks slightly complicated at first, but after expanding you're done before you know it.

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Practice Problem: Show that if $ \frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n} $, then

$ \frac{a_i}{b_i} = \frac{a_1+a_2+\cdots+a_n}{b_1+b_2+\cdots+b_n} $

for $ i = 1, 2, \ldots, n $.

Just express every a_i, b_i as (c_i a_1) / (c_i b_1) and the solution is immediate.

ReplyDeleteso ehh.. what does the rounded-E mean in the problem? and N.. where did that come from??

ReplyDeleteI get what QC's saying!!! that rarely happens...

the rounded E is for "in"

ReplyDeleteThe bold-faced N means the natural numbers, aka {1,2,3,...}.

ReplyDeleteA note to commenters: Using the less-than symbol will cut short your reply.

ReplyDeleteUnless you're really cool and use & l t ; instead =D

ReplyDelete<

ReplyDeleteHaha nice.