prove the binomial theorem?

The binomial theorem states that for any non-negative integer n and any real or complex numbers x and y,

(x + y)^n = Σ(k=0 to n) (n choose k) x^k y^(n-k)

where (n choose k) is the binomial coefficient, defined as

(n choose k) = n! / (k! (n-k)!)

where n! denotes the factorial of n, defined as the product of all positive integers less than or equal to n.

To prove the binomial theorem, we can use mathematical induction.

Base case: When n = 0, both sides of the equation are equal to 1, since (x + y)^0 = 1 and (0 choose 0) = 1. The base case is true.

Inductive step: Assume that the binomial theorem holds for some positive integer n, i.e.,

(x + y)^n = Σ(k=0 to n) (n choose k) x^k y^(n-k)

We need to show that the theorem also holds for n+1, i.e.,

(x + y)^(n+1) = Σ(k=0 to n+1) (n+1 choose k) x^k y^(n+1-k)

Using the binomial theorem for (x+y)^n, we can expand the left-hand side of the equation as

(x + y)^(n+1) = (x + y) (x + y)^n = (x + y) Σ(k=0 to n) (n choose k) x^k y^(n-k)

Expanding the product, we get

(x + y) Σ(k=0 to n) (n choose k) x^k y^(n-k) = Σ(k=0 to n) (n choose k) x^(k+1) y^(n-k) + Σ(k=0 to n) (n choose k) x^k y^(n-k+1)

To simplify these two sums, we can split them into two smaller sums by changing the index of summation:

Σ(k=0 to n) (n choose k) x^(k+1) y^(n-k) = Σ(j=1 to n+1) (n choose j-1) x^j y^(n+1-j)

Σ(k=0 to n) (n choose k) x^k y^(n-k+1) = Σ(j=0 to n) (n choose j) x^j y^(n+1-j)

Note that in the first sum, we have replaced k with j-1, and in the second sum, we have kept the index k the same but changed the limits of summation.

Now we can combine the two sums and factor out y^(n+1):

Σ(j=1 to n+1) (n choose j-1) x^j y^(n+1-j) + Σ(j=0 to n) (n choose j) x^j y^(n+1-j)

= (n choose 0) x^0 y^(n+1) + Σ(j=1 to n) [(n choose j-1) + (n choose j)] x^j y^(n+1-j) + (n choose n) x^n y^0

= Σ(j=0 to n+1) (n+1 choose j) x^j y^(n+1-j)

In the last step, we have used the identity

(n choose j-1) + (n choose j) = (n+1 choose j)

which can be proved using the formula for the binomial coefficient.

Therefore, we have shown that the binomial theorem holds for n+1. By mathematical induction, the theorem holds for all non-negative integers n.

The binomial theorem states that for any positive integer n, the expansion of (a + b)^n can be written as the sum of terms where each term is of the form C(n, k) * a^(n-k) * b^k, where C(n, k) represents the binomial coefficient.

Step 1: Let's consider the base case for n = 0.
(a + b)^0 = 1
The base case is true.

Step 2: Assume the binomial theorem is true for some positive integer k, where k >= 0.
(a + b)^k = C(k, 0) * a^(k-0) * b^0 + C(k, 1) * a^(k-1) * b^1 + ... + C(k, k-1) * a^(k-(k-1)) * b^(k-1) + C(k, k) * a^(k-k) * b^k

Step 3: We need to prove that the binomial theorem holds for n = k + 1.
(a + b)^(k+1) = (a + b)(a + b)^k
= (a + b)[C(k, 0) * a^k * b^0 + C(k, 1) * a^(k-1) * b^1 + ... + C(k, k-1) * a * b^(k-1) + C(k, k) * a^0 * b^k]

Step 4: Distribute (a + b) to each term inside the brackets.
(a + b)(a + b)^k = C(k, 0) * a^k * b^0 * (a + b) + C(k, 1) * a^(k-1) * b^1 * (a + b) + ... + C(k, k-1) * a * b^(k-1) * (a + b) + C(k, k) * a^0 * b^k * (a + b)

Step 5: Simplify each term by distributing.
(a + b)(a + b)^k = C(k, 0) * a^(k+1) * b^0 + C(k, 0) * a^k * b^1 + C(k, 1) * a^k * b^1 + C(k, 1) * a^(k-1) * b^2 + ... + C(k, k-1) * a^1 * b^k + C(k, k) * a^0 * b^(k+1)

Step 6: Combine like terms.
(a + b)(a + b)^k = [C(k, 0) + C(k, 1)] * a^(k+1) * b^0 + [C(k, 0) + C(k, 1)] * a^k * b^1 + ... + [C(k, k-1) + C(k, k)] * a^1 * b^k + C(k, k) * a^0 * b^(k+1)

Step 7: Use the definition of binomial coefficients.
(a + b)(a + b)^k = C(k+1, 0) * a^(k+1) * b^0 + C(k+1, 1) * a^k * b^1 + ... + C(k+1, k) * a^1 * b^k + C(k+1, k+1) * a^0 * b^(k+1)

Step 8: Rewrite (a + b)^(k+1) in the form of the binomial theorem.
(a + b)^(k+1) = C(k+1, 0) * a^(k+1) * b^0 + C(k+1, 1) * a^k * b^1 + ... + C(k+1, k) * a^1 * b^k + C(k+1, k+1) * a^0 * b^(k+1)

Step 9: Since the binomial theorem holds for the base case and the assumption that it holds for k, it holds for k+1 as well.

By the principle of mathematical induction, the binomial theorem is proven for any positive integer n.