Math IA Help

Godspeedyou · February 2, 2016

Hello,

For my math IA I'm investigating the relationship between Pascal's Triangle and compound interest. To do this, I need to find the ratio of successive row products, which can be seen in the equation below. I understand dividing the sequence of products with n + 1 by the sequence of products with n, but I don't understand why the sequence of products equation would be k!^(-2) and why there would be (n+1)!^(n+2) divided by n!^(n+1). For reference, the equation for sequence of products in a row of Pascal's Triangle is in the picture above. Any help would be much appreciated, thank you.

kw0573 · February 2, 2016

You've taken a long time writing this equation up, unfortunately others cannot provide help on the technical / equation level for the IA. sorry :/
However, one strategy you can try to figure the problem, is to write out the terms instead of using the product notation.

Vioh · February 3, 2016

Hello,
For my math IA I'm investigating the relationship between Pascal's Triangle and compound interest. To do this, I need to find the ratio of successive row products, which can be seen in the equation below. I understand dividing the sequence of products with n + 1 by the sequence of products with n, but I don't understand why the sequence of products equation would be k!^(-2) and why there would be (n+1)!^(n+2) divided by n!^(n+1). For reference, the equation for sequence of products in a row of Pascal's Triangle is in the picture above. Any help would be much appreciated, thank you.

Contrary to kw0573, I think this problem is way easier if you use the product notation directly. Without using the intermediate expression as done by wikipedia, I'll attempt to prove directly that:

$gif.latex? \frac{S_{n+1}}{S_n} = \frac{($

Now from the definition of , we have:

$gif.latex? \frac{S_{n+1}}{S_n} = \prod_{$

Because we are using the product notation, it's possible for us to easily separate numerators and denominators into several parts (A, B, & C).

$gif.latex? \frac{S_{n+1}}{S_n} = A \time$

Let's now calculate part A, B, & C individually. For part A, there's no k in the expression, thus all we need to do is to count how many times the multiplication is repeated. So:

$gif.latex? A = \frac{\prod_{k=0}^{n+1}$

For B, we can see that on top can be cancelled out completely by on the bottom, if and only if in the nominator and in the denominator (where 'a' is a particular value). And then all you have left is an instance of where down in the denominator. So:

$gif.latex? B = \frac{\prod_{k=0}^{n} (n-$

With similar reasoning, you should be able to derive the result for part C:

$gif.latex? C = \frac{\prod_{k=0}^{n} k!}$

Combining all the parts, we get:

$gif.latex? \frac{S_{n+1}}{S_n} = A \time$

Due to the complexity of all the equations, I might have made some typos here and there. So if anybody spots any, please tell me so that I can fix it.

Edited February 3, 2016 by Vioh

Godspeedyou · February 3, 2016

Hello,
For my math IA I'm investigating the relationship between Pascal's Triangle and compound interest. To do this, I need to find the ratio of successive row products, which can be seen in the equation below. I understand dividing the sequence of products with n + 1 by the sequence of products with n, but I don't understand why the sequence of products equation would be k!^(-2) and why there would be (n+1)!^(n+2) divided by n!^(n+1). For reference, the equation for sequence of products in a row of Pascal's Triangle is in the picture above. Any help would be much appreciated, thank you.
Contrary to kw0573, I think this problem is way easier if you use the product notation directly. Without using the intermediate expression as done by wikipedia, I'll attempt to prove directly that:
$gif.latex? \frac{S_{n+1}}{S_n} = \frac{($
Now from the definition of , we have:
$gif.latex? \frac{S_{n+1}}{S_n} = \prod_{$
Because we are using the product notation, it's possible for us to easily separate numerators and denominators into several parts (A, B, & C).
$gif.latex? \frac{S_{n+1}}{S_n} = A \time$
Let's now calculate part A, B, & C individually. For part A, there's no k in the expression, thus all we need to do is to count how many times the multiplication is repeated. So:
$gif.latex? A = \frac{\prod_{k=0}^{n+1}$
For B, we can see that on top can be cancelled out completely by on the bottom, if and only if in the nominator and in the denominator (where 'a' is a particular value). And then all you have left is an instance of where down in the denominator. So:
$gif.latex? B = \frac{\prod_{k=0}^{n} (n-$
With similar reasoning, you should be able to derive the result for part C:
$gif.latex? C = \frac{\prod_{k=0}^{n} k!}$
Combining all the parts, we get:
$gif.latex? \frac{S_{n+1}}{S_n} = A \time$
Due to the complexity of all the equations, I might have made some typos here and there. So if anybody spots any, please tell me so that I can fix it.