# Diminished Radix Complement (geometric pattern)

#### deersfeet

So I'm reading about Radix Complements on Wikipedia, and there is a geometric pattern used to prove the following statement:
The radix complement is most easily obtained by adding 1 to the diminished radix complement, which is (bn-1)-y. Since bn-1 is the digit b-1 repeated n times because
bn-1 = bn-1n = (b-1)(bn-1+bn-2...+b+1) = (b-1)bn-1+ (b-1)bn-2 +...+b-1

I'm trying to understand the third step in this pattern. I understand that an intermediary step would be (b-1)n but why does this translate into (b-1)(bn-1+bn-2...+b+1)?

Thanks,
deersfeet

#### Archie

Try expanding the brackets in the third step. You should get back to the second step (with no intermediates).

#### kalyanram

I'm trying to understand the third step in this pattern. I understand that an intermediary step would be (b-1)n but why does this translate into (b-1)(bn-1+bn-2...+b+1)?
You are getting confused between two things here
First establish the following
1. Using the binomial expansion $$\displaystyle (b-1)^n = \sum^n_{i=0} (-1)^n C^n_i b^{n-i}$$, so $$\displaystyle b^n-1^n \neq (b-1)^n$$, where $$\displaystyle C^n_i = \frac{n!}{(n-i)!. i!}$$.

2.Say $$\displaystyle S_n = b^{n-1}+b^{n-2}+....b^2+b+1$$ (Eq 1)

$$\displaystyle b.S_n = b^{n}+b^{n-1}+....b^2+b$$ (Eq 2)

Subtracting (Eq 1) from (Eq 2) we have

$$\displaystyle b.S_n - S_n= b^{n}-1 \implies (b-1)S_n = b^n-1$$ as already stated.

Hope this helped.
~Kalyan.

#### deersfeet

Hi Kalyan,
Thanks for responding! Your final answer makes some sense to me but I don't understand some of the notation. What do b.Sn and (n-1)!.i! mean?

#### kalyanram

What do b.Sn and (n-1)!.i! mean?
$$\displaystyle b.S_n$$ means $$\displaystyle S_n$$ is multiplied by $$\displaystyle b$$ and $$\displaystyle n!$$ is read as "n factorial" for some positive integer $$\displaystyle n$$.

You may look into binomial theorem and summation of geometric series to have a better understanding into this. Hope it helps.

~K

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