Interesting problem

Valentas

New member
Joined
Dec 19, 2010
Messages
34
One after another number 2 degrees wrote so that in the first line there is one number, in the second 2, in the third 3, in the fourth 4 and so on...

1
2 4
8 16 32
64 128 256 512
... ... ... .... ....

Show that this in this table's row n all numbers product is calculated by formulae

\(\displaystyle S(n) = 2^(\frac{n^3-n}{2})\)

It is 2 powered by \(\displaystyle \frac{n^3-n}{2} \)
 
Why is that hard?

The first element in each row is 2^0, 2^1, 2^3, 2^6, 2^10... You can codify that.

Each element is each row is divisible by the first element. Factoring this out leaves a finite geometric progression starting at '1'.

Let's see what you get.
 
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