**
A formula for the total
number of biological ancestors for a given number of generations back in history**

We seek
to derive a formula for an individual’s total number of biological ancestors for
a given number of generations back. Beginning, we observe that at one
generation back, there are two biological parents. Each biological parent has
two biological parents which are the original individual’s biological
grandparents. With this in mind, it is intuitively obvious that there are 2^{n}
biological ancestors of the original individual at each generation where n is
the number of generations back for that particular generation.

However,
the total number of biological ancestors is the sum of the number of biological
ancestors at each generation level down through the original individual’s
biological parents. So, the total number of biological ancestors is of the form
2^{1} + 2^{2} + … + 2^{n}, where n is the number of
generations back. If one examines the value of this expression for the first
several positive integer values of n, it begins to appear as if the value
equates to the expression 2^{n+1 }- 2.

Can we prove the following?

2^{n+1}
- 2 = 2^{1} + 2^{2} + … + 2^{n}, where n is a positive
integer

We begin by adding 2 to both sides of the equation

2^{n+1}
= 2 + 2^{1} + 2^{2} + … + 2^{n}

Now we divide both sides by 2

2^{n}
= 1 + 2^{0} + 2^{1} + … + 2^{n-1}

Now we subtract 1 from both sides

2^{n}
– 1 = 2^{0} + 2^{1} + … + 2^{n-1}

An inductive proof of the above equation does exist for all positive integers n, and I will now attempt to explain it as fully as I am able.

To inductively prove an equation in the variable n, one must do two things:

1) One must show that the equation is true for the lowest value of n for which we expect the equation to be true. In this case, we don’t even care whether the equation is true for values of n less than 1 or for any values of n which are not integers. Therefore, for this step we will establish that the equation is true for n=1.

2)
After step 1,
we must then show that ** if** the equation is true for some specific
integer value of n (we will call this specific value k),

Let’s restate our altered version of the equation that I said we could prove:

2^{n}
– 1 = 2^{0} + 2^{1} + … + 2^{n-1}

For n=1,

2^{1}
- 1= 2^{1-1}

Before
we evaluate this and see if the two sides are equal, I want to address any
confusion which may have arisen. I admit that the equation

2^{n} – 1 = 2^{0} + 2^{1} + … + 2^{n-1} seems to
imply that the right hand side will be the sum of at least three terms. For
n=3, it would consist of three terms. But we’ve got n=1, so let’s consider what
the expression 2^{0} + 2^{1} + … + 2^{n-1} is really
saying. Translated from the language of mathematics into the English language
it says, “the sum of every nonnegative integer power of 2 up to n-1”. Since we
are dealing with n=1, then n-1=0. So we read that again as, “the sum of every
nonnegative integer power of 2 up to 0”. This means from 2^{0} all the
way up to 2^{0}. That means just a simple 2^{0}. Returning to
our arithmetic, we write the next step as

2 – 1 =
2^{0}

or

1 = 1

Step 1 of the inductive proof has been a success.

Turning our attention to step 2, consider the following sentence:

“If there is an elephant in the room, the floor has undergone excessive stress.”

We can prove this statement by researching the weight of elephants, and by determining the stress which would be considered excessive for the floor. If we do find that the weight of an elephant would place excessive stress on the floor, then the statement is true, even if there never has been and never will be an elephant in the room. Metaphorically speaking, we have determined that A implies B. We never said that A was true. We never said that B was true. We only said, “If A, then B.” The word “if” allows our sentence to be true, even though it may be hypothetical. Nevertheless, if we ever could prove A then we’ve proved B, keeping in mind that earlier we did prove that A implies B.

We have an equation

2^{n}
– 1 = 2^{0} + 2^{1} + … + 2^{n-1}

We haven’t proved the equation yet, but the “equation” as written may have implications. It may be able to imply something for us.

Referring back to our preview of step 2, we proceed by rewriting the above equation for n=k and n=k+1:

For n=k

2^{k}
– 1 = 2^{0} + 2^{1} + … + 2^{k-1}

and for n=k+1

2^{k+1}
– 1 = 2^{0} + 2^{1} + … + 2^{k-1}
+ 2^{k}

Note
that the terms in red above are, according to our equation for case n=k, equal
to 2^{k} – 1. So we may write

2^{k+1}
– 1 = 2^{k} – 1 + 2^{k}

2^{k+1}
– 1 = 2(2^{k}) – 1

2^{k+1}
– 1 = 2^{k+1} – 1

Now we have shown that case n=k implies case n=k+1. Earlier in step 1 we showed that the equation is true for n=1, so it stands to reason that the equation of case n=k is true for k=1. And if that equation is true for n=k=1, it is also true for n=k+1=2. Furthermore, the fact that case n=k implies case n=k+1 means that case n=k+1 implies case n=k+1+1, etc. Step 2 is a success. We have proved that the original equation is true for all positive integers. The total number of biological ancestors for n generations back in history is given by the expression

2^{n+1}
- 2