# Thread: A Product Series Equation

1. ## A Product Series Equation

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

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

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

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

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

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

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

In the numerators of the fractions on the RHS, all but the last three, $\displaystyle x-2$, $\displaystyle x-1$, and $\displaystyle x$, 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

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

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