Since you guys seem to be fans of the binomial theorem, I cooked up the proof using mathematical induction for you guys
Proof of the Binomial Theorem:
Step 1: Base Case
For n = 1, the binomial theorem states:
(x + y)^1 = C(1, 0) x^1 y^0 + C(1, 1) x^0 y^1
which simplifies to:
x + y = x + y
This confirms that the base case holds true.
Step 2: Inductive Hypothesis
Assume that the binomial theorem holds true for some positive integer k:
(x + y)^k = C(k, 0) x^k y^0 + C(k, 1) x^(k-1) y^1 + C(k, 2) x^(k-2) y^2 + ... + C(k, k-1) x^1 y^(k-1) + C(k, k) x^0 y^k
Step 3: Inductive Step
We need to prove that the binomial theorem also holds true for k + 1. So, let's consider (x + y)^(k+1):
(x + y)^(k+1) = (x + y) * (x + y)^k
Using the distributive property, we can expand this as:
(x + y)^(k+1) = (x + y) (C(k, 0) x^k y^0 + C(k, 1) x^(k-1) y^1 + C(k, 2) x^(k-2) y^2 + ... + C(k, k-1) x^1 y^(k-1) + C(k, k) x^0 * y^k)
Expanding further and simplifying, we get:
(x + y)^(k+1) = C(k, 0) x^(k+1) y^0 + C(k, 1) x^k y^1 + C(k, 2) x^(k-1) y^2 + ... + C(k, k-1) x^2 y^(k-1) + C(k, k) x^1 y^k
- C(k, 0) x^k y^1 + C(k, 1) x^(k-1) y^2 + C(k, 2) x^(k-2) y^3 + ... + C(k, k-1) x^0 y^k + C(k, k) x^0 y^(k+1)
Using the binomial coefficient identity C(n, r) + C(n, r+1) = C(n+1, r+1), we can simplify the expression further:
(x + y)^(k+1) = (x + y)^k + C(k, 0) x^k y^1 + C(k, 1) x^(k-1) y^2 + C(k, 2) x^(k-2) y^3 + ... + C(k, k-1) x^0 y^k + C(k, k) x^0 y^(k+1)
This can be rewritten as:
(x + y)^(k+1) = (x + y)^k + [C(k, 0) x^k y^1 + C(k, 1) x^(k-1) y^2 + C(k, 2) x^(k-2) y^3 + ... + C(k, k-1) x^0 y^k] + C(k, k) x^0 y^(k+1)
Notice that the term in the brackets matches the binomial theorem for k, so we can rewrite it as (x + y)^k. This simplifies the expression to:
(x + y)^(k+1) = (x + y)^k + (x + y) C(k, k) x^0 * y^(k+1)
Simplifying further, we have:
(x + y)^(k+1) = (x + y)^k + C(k+1, k+1) x^0 y^(k+1)
Since C(k+1, k+1) = 1, the expression becomes:
(x + y)^(k+1) = (x + y)^k + x^0 * y^(k+1)
This confirms that the binomial theorem holds true for k + 1.
Step 4: Conclusion
Since the base case holds true, and assuming that the theorem holds for k implies it holds for k + 1, we can conclude that the binomial theorem holds true for all positive integers.
Here is the link to the Pythagorean Theorem proof if you want to see more
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