A Product Series Equation

**Feryll** · November 25th 2009, 06:29 PM

Before I get started, I am not sure if this should be in pre-calculus or here, so please forgive me if it is in the wrong place.
Anyhow, I came upon this supposed equality in Picture Num 1.
I originally found it in Picture Number 2.
Could someone please explain why the equality in Picture Number One exists? Like what law states that it is true? Thanks
P.S. If you can't see, the expression on the right of the sigma in Picture Number 2 is L^4

**Grandad** · November 26th 2009, 07:37 AM

Hello Feryll

Welcome to Math Help Forum!

Originally Posted by Feryll

Before I get started, I am not sure if this should be in pre-calculus or here, so please forgive me if it is in the wrong place.
Anyhow, I came upon this supposed equality in Picture Num 1.
I originally found it in Picture Number 2.
Could someone please explain why the equality in Picture Number One exists? Like what law states that it is true? Thanks
P.S. If you can't see, the expression on the right of the sigma in Picture Number 2 is L^4

Are you familiar with Proof by Induction? This is quite a straightforward appliction of this method.

Suppose that $P(x)$ is the propositional function $\prod_{i=3}^x\left(1+\frac{3}{i-3}\right)=\frac{x^3-3x^2+2x}{6}=\frac{x(x-1)(x-2)}{6}$

Then $P(x) \Rightarrow \prod_{i=3}^{x+1}\left(1+\frac{3}{i-3}\right)=\frac{x(x-1)(x-2)}{6}\cdot\left(1+\frac{3}{x-2}\right)$

$=\frac{x(x-1)(x-2)}{6}\cdot\left(\frac{x+1}{x-2}\right)$

$=\frac{[x+1]([x+1]-1)([x+1]-2)}{6}$

$\Rightarrow P(x+1)$

Now $P(4)$ is $\prod_{i=4}^4\left(1+\frac{3}{i-3}\right)=\frac{(4)(3)(2)}{6}$ , which is true.

So $P(x)$ is true for all $x \ge 4$ .

Grandad

**awkward** · November 26th 2009, 07:48 AM

Originally Posted by Feryll

Before I get started, I am not sure if this should be in pre-calculus or here, so please forgive me if it is in the wrong place.
Anyhow, I came upon this supposed equality in Picture Num 1.
I originally found it in Picture Number 2.
Could someone please explain why the equality in Picture Number One exists? Like what law states that it is true? Thanks
P.S. If you can't see, the expression on the right of the sigma in Picture Number 2 is L^4

$1+ \frac{3}{i-3} = \frac{i}{i-3}$ ,
so
$\prod_{i=4}^x \left(1 + \frac{3}{i-3} \right) = \frac{4}{1} \cdot \frac{5}{2} \cdot \frac{6}{3} \cdot \frac{7}{4} \cdots \frac{x-3}{x-6} \cdot \frac{x-2}{x-5} \cdot \frac{x-1}{x-4} \cdot \frac{x}{x-3}$

In the numerators of the fractions on the RHS, all but the last three, $x-2$ , $x-1$ , and $x$ , cancel with the denominators of the preceding fractions. In the numerators, all but the first three, 1, 2, and 3, cancel with the denominators. So

$\prod_{i=4}^x \left(1 + \frac{3}{i-3} \right) = \frac{(x-2)(x-1)(x)}{1\cdot 2 \cdot 3} = (1/6) \; (x^3 - 3x^2 + 2x)$

[Edit]Beaten to the punch by Grandad! But my solution is different anyway.[/Edit]

Thread: A Product Series Equation

LinkBack

Thread Tools

Display

A Product Series Equation

Similar Math Help Forum Discussions

What's next in the series: Sum, Product, ...

product of 2 series

sum of the product of convergent series

Product of Series

Rearrange product of two series

Search Tags