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How would I write this as an integer or a fraction in lowest terms?

$(1-\frac12)(1+\frac 12)(1-\frac13)(1+\frac13)(1-\frac14)(1+\frac14).....(1-\frac1{99})(1+\frac1{99})$

I really need to understand where to start and the process if anyone can help me.

azarel
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SNS
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2 Answers2

9

Rewrite it instead as

$$\color{Green}{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdots\left(1-\frac{1}{98}\right)\left(1-\frac{1}{99}\right)}\times \color{Blue}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{98}\right)\left(1+\frac{1}{99}\right)}$$

Evaluate left and right parts separately. Hint: Look at how numerator and denominator cancel.

For example,

$$\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)=\frac{\color{DarkOrange}1}{\color{Red}2}\cdot\frac{\color{Red}2}{\color{Green}3}\cdot\frac{\color{Green}3}{\color{Blue}4}\cdot\frac{\color{Blue}4}{\color{DarkOrange}5}=\frac{1}{5}.$$

anon
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  • So are you saying that in (1-1/2) and (1+1/2) that the 1/2 cancels each other out? – SNS Feb 28 '12 at 00:36
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    @SNS: No, I am not saying anything about 1-1/2 in relation to 1+1/2. I am telling you to separate those factors so you can deal with the minus factors independently of the plus factors. In other words, we are rearranging $a_1b_1\cdots a_n b_n$ to $a_1\cdots a_n b_1\cdots b_n$. – anon Feb 28 '12 at 00:39
  • Ok, so in the minus factors it would leave 1/99! – SNS Feb 28 '12 at 00:51
  • @SNS: Exactly! Now experiment with the plus factors and see if you can put it all together... – anon Feb 28 '12 at 00:54
  • And in the plus factors it would leave 100/2=50 – SNS Feb 28 '12 at 00:57
  • So the answer would be 50/99???? – SNS Feb 28 '12 at 00:57
  • @SNS: That is the correct answer, congratulations! – anon Feb 28 '12 at 01:00
  • Thank you so much!! I feel a weight lifted. I have been struggling with this for 2 weeks! – SNS Feb 28 '12 at 01:01
8

Hint: telescopy, note how the adjacent like-colored terms all cancel out of the products below

$$\rm \left(1-\frac{1}2\right)\left(1-\frac{1}3\right)\cdots \left(1-\frac{1}n\right)\ =\ \frac{1}{\color{red}2} \frac{\color{red}2}{\color{green}3} \frac{\color{green}3}{\color{blue}4} \frac{\color{blue} 4}{}\: \cdots\: \frac{}{\color{brown}{n-1}}\frac{\color{brown}{n-1}}n\ =\ \frac{1}n $$

$$\rm \left(1+\frac{1}2\right)\left(1+\frac{1}3\right)\cdots \left(1+\frac{1}n\right)\ =\ \frac{\color{red}3}2 \frac{\color{green}4}{\color{red}3} \frac{\color{blue}5}{\color{green}4}\frac{}{\color{blue}5}\: \cdots\: \frac{\color{brown} n}{}\frac{n+1}{\color{brown}n}\ =\ \frac{n+1}2$$

Hence $$\rm \prod_{k\:=\:2}^n\: \left( 1-\frac{1}{k^2}\right)\ =\ \frac{n+1}{2\:n} $$

Follow the above link for many more examples of telescopic proofs (additive and multiplicative).

Community
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Bill Dubuque
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  • Thanks for laying it out like this, I was able to see what numbers canceled each other out! – SNS Feb 28 '12 at 01:00