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$(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{n})$=$(\frac{1}{n})$ So I'm trying to make it equal.

so $n$ is equal or greater than 2. When i substitute $2$ to n, $(1-\frac{1}{2})$ = $(\frac{1}{2})$

then
$(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{k})$=$(\frac{1}{k})$

$(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{k})+(1-\frac{1}{K+1})$=$(\frac{1}{k+1})$

I can't make it equal I got an answer of

$\frac{1}{k}+\frac{k}{k+1}=\frac{1}{k+1}$ I don't know if N greater than or equal to 2 has something to do with it. The usual problems that I solve just prove n is equal to 1. I did the algebra and still can't make my LHS equal to $\frac{1}{k+1}$

Mickey
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  • How can i make it equal. – Mickey Feb 18 '15 at 09:52
  • @yashg how come it says here http://www.mathsisfun.com/algebra/mathematical-induction.html in the solutions that i need to add? – Mickey Feb 18 '15 at 10:01
  • In the induction step, you assume that some statement/formula/equation/etc holds for some $n$, then you are to prove that it holds for $n+1$. That is, if $P(n)$ holds then $P(n+1)$ holds. In your case $P(n)$ is $$\prod\limits_{k=2}^n\left(1-\frac{1}{k}\right) = \frac{1}{n}.$$ What is $P(n+1)$? – Eff Feb 18 '15 at 10:07
  • @Eff $\frac{1+n}{n}$? – Mickey Feb 18 '15 at 10:12
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    @Mickey What? No, it would be $$\prod\limits_{k=2}^{n+1}\left(1-\frac{1}{k}\right) = \frac{1}{n+1},$$ which is the same as $$\left(1-\frac{1}{n+1}\right)\prod\limits_{k=2}^{n}\left(1-\frac{1}{k}\right) = \frac{1}{n+1}. $$ – Eff Feb 18 '15 at 10:15
  • @yashg I'm actually doing an advance study and my girl teacher is so lazy to teach so i consider to to advance studying sorry. I got it now I didn't know they are different for sum and product thank you so much i just learn from youtube videos – Mickey Feb 18 '15 at 10:24

1 Answers1

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As yashg mentions, you are adding and not multiplying. This might help:

$$\left(\prod_{k=2}^n\left(1-\frac1k\right)\right)\left(1-\frac1{n+1}\right)=\prod_{k=2}^{n+1}\left(1-\frac1k\right)$$

And

$$\begin{align}\left(\frac1n\right)\left(1-\frac1{n+1}\right)&=\frac1n-\frac1{n(n+1)}\\ &=\frac{n+1}{n(n+1)}-\frac1{n(n+1)}\\& =\frac{n+1-1}{n(n+1)}\\ &=\frac{1}{n+1}\end{align}$$

Regret
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  • how come here http://www.mathsisfun.com/algebra/mathematical-induction.html it says add... Is there like a scenario where i need to multiply or just add? – Mickey Feb 18 '15 at 09:59
  • @Mickey: Here is a link to a question about how induction works, you may find it helpful. – Regret Feb 18 '15 at 10:05
  • @Mickey: In this specific question you have asked, you are dealing with a product, not a sum. To get from the case $n$ to the case $n+1$, you should multiply by the $n+1$-th term. – Regret Feb 18 '15 at 10:07
  • How do i know if I need to deal with a product? I'm still reading the link you gave me – Mickey Feb 18 '15 at 10:08
  • @Mickey: Your equation has many terms multiplied together on the left side (a product). – Regret Feb 18 '15 at 10:16
  • NOW I GET IT! Sorry I was just learning from youtube, but a lot of thanks my friend sorry again. – Mickey Feb 18 '15 at 10:26