# induction

• Sep 11th 2007, 07:17 PM
PvtBillPilgrim
induction
I have the cartesian product of this:

1 - (1/(k^2)) from k = 2 to n.

I picked random values of n like n = 2, 3, 4.

With n = 2, I have (1 - (1/4)).
With n = 3, I have (1 - (1/4)) * (1 - (1/9)).
With n = 4, I have (1 - (1/4)) * (1 - (1/9)) * (1 - (1/16)).
etc.

How do I go about finding a formula for this for all n?

I know that 1 - (1/(k^2)) = ((k + 1)(k - 1))/k^2.

Thanks in advance for any help.
• Sep 11th 2007, 07:41 PM
Soroban
Hello, PvtBillPilgrim!

Quote:

I have the cartesian product of this: . $1 - \frac{1}{k^2}\;\text{ for }k \,= \,2\text{ to }n.$

How do I go about finding a formula for this?

You should have examined those cases more closely . . .

$\begin{array}{cccc}n = 2: & 1 - \frac{1}{2^2} & = & \frac{3}{4} \\ \\

n = 3: & \left(\frac{3}{4}\right)\left(1-\frac{1}{3^2}\right) & = & \frac{4}{6} \\ \\

n = 4: & \left(\frac{4}{6}\right)\left(1-\frac{1}{4^2}\right) & = & \frac{5}{8} \\ \\

n=5: & \left(\frac{5}{8}\right)\left(1-\frac{1}{5^2}\right) & = & \frac{6}{10} \end{array}$

It appears that the general form is: . $P_n \;=\;\frac{n+1}{2n}$

The next task is to prove it . . .