Proof that ∑2^(n-1) = 2^n - 1 with Mathematical Induction

In this video I demonstrate that the equation 1 + 2 + 2² + 2³ + ... + 2^n-1 = 2ⁿ - 1 for all positive integers using mathematical induction.

The first step in this process is the basis test, which is to see if this statement is true for the first natural number n = 1. With n = 1, we get on the left hand side...

2^0 = 1

And on the right hand side, we get...

2^1 - 1 = 2 - 1 = 1

So LHS = RHS and this statement passes the basis test.

The next step is the inductive step, where we prove that the statement holds true for any natural number.

First, let's assume the statement is true for n = k. Thus...

1 + 2 + 2^2 + 2^3 + ... + 2^{k-1} = 2^k - 1

If it is true for n = k, then it should also be true for when n = k + 1. Let's put this to the test! So when n = k + 1, the left hand side equals...

1 + 2 + 2^2 + 2^3 + ... + 2^{k-1} + 2^k

On the right hand side, we have...

2^{k+1} - 1

Now, up to the kth term, this is equal to...

2^k - 1

Thus the left hand side equals...

2^k - 1 + 2^k = 2\cdot2^k - 1 = 2^{k+1} = RHS

And therefore, we conclude that this statement is true for any natural number n