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I have the following, which I have arrived at after a series of calculations.

$\prod_1^n 6.\bigg\{2 - \frac{5}{6}\Big(n\Big)\bigg\}.$

My problem is my maths is a little rusty...Could someone explain to me how to simplify this?

I'm just not entirely sure what repercussions the curly braces have around the

EDIT: I have another post discussing another allowed method using geometric series also, but again couldn't arrive at the answer...

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    Is that $2-\frac{5}{6n}$ or $2-\frac{5}{6}n$? Parentheses, please. The curly brackets are just one kind of parentheses.2011-04-13
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    sorry , it's $2-\frac{5}{6}n.$2011-04-13
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    It would help if we had more background; such as how you arrived at this formula. We can then help you get the basics and make sure that you get everything right. Your formulas appear to describe two different things. I get the impression that you are trying to describe part of a geometric series, though. Once we get this right, it will be easy to fill in the blanks and get a simple answer. So some background as to how you got here would really help. It would help if we could see the terms listed out, for instance, like 2(5/6), $2(5/6)^2$, ... $2(5/6)^6$.2011-04-13
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    This is a follow up to this post http://math.stackexchange.com/questions/32726/simple-maths-question-capital-sigma-pi2011-04-13
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    Indeed. And of http://math.stackexchange.com/questions/32598 and of http://math.stackexchange.com/questions/32549 as well. Rather than posting on this site FOUR highly related questions in less than 24 hours, it might be more productive to *digest* the answers to ONE of them or to ask some further explanations on these. Just my two cents.2011-04-13
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    I see why user9492 did this, though - I'm working on http://math.stackexchange.com/questions/32598 to provide another answer. I hope that this will resolve the issue. I think that the OP would benefit from multiple answers.2011-04-13
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    @Matt: Care to share your insight on why?2011-04-13
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    I think that user9492 is trying very hard to learn this problem, as evidenced by all of the work shown and the interactions here. So more answers should probably provide a more thorough understanding. It seems hard to touch upon everything in one answer, I guess.2011-04-13

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$\prod_1^n$ means product as something runs from $1$ to $n$ - but what something? Usually one sees, say, $\prod_{k=1}^n$, then you know it's $k$ that goes from $1$ to $n$, but if you leave out the $k$ then, unless there's some context, the product symbol is ambiguous.

Usually, if it's $\prod_{k=1}^n$, then what follows has $k$ in it, not $n$ - but in what you've written, there's an $n$.

These things make it difficult to know what to make of your formula.

On top of that, it's not clear whether $5/6n$ means $(5/6)n$ or $5\over 6n$.

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    Well my formula says the product from 1 to n (n being some arbitrary number greater than 0). The 6 is a constant..I think I have just talked myself through the solution...would the above then mean (6^n) * $2-\frac{5}{6}n .$?2011-04-13
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    @user9492: As you have written it, you would just have $6^n(2-\frac{5n}{6})^n=(12-5n)^n$2011-04-13
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    I have been trying to do this question for ages and my lecture notes aren't sufficient to solve the problem, even though the lecturer says they are. I am trying to solve $f(n) = 6f(n-1) + 5.$ The base case $f(0) = 2.$ Apparently the answer should be $.(6^n) + 1.$ I have no idea how to get to this using the above formula.2011-04-13
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Assuming, like Gerry says, that what you want is $\prod_{k=1}^n 6(2 - \frac{5k}{6})$, you can multiply the $6$ in and get $\prod_{k=1}^n (12 - 5k)=7\cdot 2\cdot -3 \cdot -8 \ldots (12-5n)$ but I don't see a convenient closed form. With the negative signs it bounces around zero, and with the multiplies it bounces a lot.

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    I don't think your interpretation is the one intended. The question is motivated by [this question](http://math.stackexchange.com/questions/32726/simple-maths-question-capital-sigma-pi) where there is a formula I can't really make sense of (I didn't try too hard, but still). But, of course, +1 nevertheless.2011-04-13
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To quote user 9492, "I have been trying to do this question for ages and my lecture notes aren't sufficient to solve the problem, even though the lecturer says they are. I am trying to solve f(n)=6f(n−1)+5. The base case f(0)=2. Apparently the answer should be .(6n)+1. I have no idea how to get to this using the above formula".

Sometimes the best way to solve a problem is just by directly looking at it and attacking the problem without any complicated formulas.

Let $a_n = 6a_{n-1} + 5$. Then:

$a_1 = 6a_0 + 5 = 6a_1 + 5, \therefore a_2 = 6(6a_0 + 5)+5 = 6^{2} a_0 + 6(5) + 5$, $a_3 = 6a_2 + 5 = 6(6^{2} a_0 + 6(5) + 5) + 5 = $ $6^{3}a_0 + 6^2 (5) + 6(5) + 5 = 6^3 a_0 + 5(1+6+6^2)$.

So in general, $a_n = 6^{n} a_0 + 5\frac{1-6^{n}}{1-6}$. As $a_0 = 2$ according to you, we have that

$a_n = 2*6^{n} - 1 + 6^{n} = 3*6^n -1$. How is it that given the recurrence relation you get $6^n +1$?

If $f(0) = 2$ according to you, then $f(1) = (6\times 2) + 5 = 17$. This is clearly not equal to $6^1 + 1$ so there's something wrong with what you have stated

If on the other hand you look at the formula for $a_n$ I have given you (sorry for switching between $a_n's$ and $f(n)'s$, $a_1 = 3*6^1 - 1 = 17$ which is consistent with the given recurrence relation and initial value of $a_0 = 2$.

Ben

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    @user9492 Looking at the previous thread, I realise how you got $6^n +1$. It's because of a different recurrence relation! Please be careful of what you type and in your comment above you talked of $f(n) = f(n-1)+5$ and not -5 as in the previous thread, please type everything carefully one minus sign is enough to change everything!2011-04-13
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    Sorry for that - I've spent so bloody long on this that I can;t even concentrate. I HAVE to use the formula above or use the method used in the linked thread. Its for an exam so I cannot prove it the way you have2011-04-13
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As an answer that summarizes the method described in user9492's questions:

The general procedure for your recurrences is given here. This link goes into great depth on the method you're learning about.

Also, the formula above can be found on page 27. This should help make everything more clear