0
$\begingroup$

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...

  • 0
    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

4 Answers 4

2

$\prod_1^n$ 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 $5\over 6n$.

  • 0
    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
1

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.

  • 1
    I don't think your interpret$a$tion is the one intended. The question is motivated by [this question](http://math.stacke$x$ch$a$nge.com/questions/32726/simple-m$a$ths-question-c$a$pit$a$l-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
1

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

  • 0
    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
0

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