prove the binomial theorem
The Binomial Theorem states that:
(x + y)^n = ∑_(k=0)^n (n choose k) x^(n-k) y^k
where (n choose k) = n!/k!(n-k)!
We can prove this result using mathematical induction.
Base case: n = 0
(x + y)^0 = 1
The formula holds true for n = 0 since (n choose k) = 0 for all k > n or k < 0. Thus the sum collapses to only the term k = 0, which is 1.
Inductive step: Assume that the formula holds true for some positive integer n.
(x + y)^n = ∑_(k=0)^n (n choose k) x^(n-k) y^k
We want to show that the formula holds true for n + 1.
(x + y)^(n+1) = (x + y)(x+y)^n = (x + y)∑_(k=0)^n (n choose k) x^(n-k) y^k
Expanding the product using distributive property:
= ∑_(k=0)^n (n choose k) x^(n-k+1) y^k + ∑_(k=0)^n (n choose k) x^(n-k) y^(k+1)
Notice that in the first sum, we can substitute j = k-1 and obtain:
∑_(j=-1)^(n-1) (n choose j+1) x^(n-j) y^(j+1)
But (n choose j+1) = (n choose j) (n-j)/(j+1) and therefore we can re-write it as:
∑_(j=-1)^(n-1) (n choose j) x^(n-j) y^(j+1) (n-j)/(j+1)
Using the relation that (n choose k) = 0 for k < 0 or k > n, we can simplify the range of summation:
(x + y)^(n+1) = ∑_(k=0)^n (n choose k) x^(n-k+1) y^k + ∑_(k=0)^n (n choose k) x^(n-k) y^(k+1)
= (n choose n) x^(n+1) + ∑_(k=0)^n (n choose k) x^(n-k+1) y^k + ∑_(k=1)^(n+1) (n choose k-1) x^(n-k+1) y^k
= x^(n+1) + ∑_(k=1)^n [(n choose k) + (n choose k-1)] x^(n-k+1) y^k + (n choose n) y^(n+1)
The sum in the middle is the same as:
∑_(k=1)^n (n+1 choose k) x^(n-k+1) y^k
Therefore:
(x + y)^(n+1) = x^(n+1) + ∑_(k=1)^n (n+1 choose k) x^(n-k+1) y^k + y^(n+1)
Which proves the formula for n+1.
Therefore, we have proven the Binomial theorem by mathematical induction.
The binomial theorem states that for any positive integer n, and any real numbers a and b:
(a + b)^n = C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n-1) * a^1 * b^(n-1) + C(n,n) * a^0 * b^n
where C(n,k) is the binomial coefficient, given by:
C(n,k) = n! / (k! * (n-k)!)
To prove the binomial theorem, we can use mathematical induction.
Step 1: Base case
For n = 0, the equation becomes:
(a + b)^0 = C(0,0) * a^0 * b^0
1 = 1 * 1 * 1
The base case holds true.
Step 2: Assume the equation holds true for n = k, where k is a positive integer.
Assume that:
(a + b)^k = C(k,0) * a^k * b^0 + C(k,1) * a^(k-1) * b^1 + C(k,2) * a^(k-2) * b^2 + ... + C(k,k-1) * a^1 * b^(k-1) + C(k,k) * a^0 * b^k
Step 3: Prove the equation holds true for n = k+1.
(a + b)^(k+1) = (a + b) * (a + b)^k
Using the assumption from Step 2:
(a + b) * (a + b)^k = (a + b) * [C(k,0) * a^k * b^0 + C(k,1) * a^(k-1) * b^1 + C(k,2) * a^(k-2) * b^2 + ... + C(k,k-1) * a^1 * b^(k-1) + C(k,k) * a^0 * b^k]
Expanding the expression:
(a + b) * [C(k,0) * a^k * b^0 + C(k,1) * a^(k-1) * b^1 + C(k,2) * a^(k-2) * b^2 + ... + C(k,k-1) * a^1 * b^(k-1) + C(k,k) * a^0 * b^k]
= [C(k,0) * a^(k+1) * b^0 + C(k,1) * a^k * b^1 + C(k,2) * a^(k-1) * b^2 + ... + C(k,k-1) * a^2 * b^(k-1) + C(k,k) * a^1 * b^k]
+ [C(k,0) * a^k * b^1 + C(k,1) * a^(k-1) * b^2 + C(k,2) * a^(k-2) * b^3 + ... + C(k,k-2) * a^2 * b^(k) + C(k,k-1) * a^1 * b^(k+1)]
= C(k,0) * a^(k+1) * b^0 + [C(k,1) * a^k * b^1 + C(k,1) * a^k * b^0] + [C(k,2) * a^(k-1) * b^2 + C(k,2) * a^(k-1) * b^1] + ...
+ [C(k,k-1) * a^2 * b^(k-1) + C(k,k-1) * a^2 * b^(k-2)] + C(k,k) * a^1 * b^k + C(k,k) * a^0 * b^(k+1)
By rearranging the terms, we get:
= C(k,0) * a^(k+1) * b^0 + [C(k,1) * a^k * b^1 + C(k,0) * a^k * b^0] + [C(k,2) * a^(k-1) * b^2 + C(k,1) * a^(k-1) * b^1] + ...
+ [C(k,k-1) * a^2 * b^(k-1) + C(k,k-2) * a^2 * b^(k-2)] + [C(k,k) * a^1 * b^k + C(k,k-1) * a^1 * b^(k+1)] + C(k,k) * a^0 * b^(k+1)
= C(k+1,0) * a^(k+1) * b^0 + C(k+1,1) * a^k * b^1 + C(k+1,2) * a^(k-1) * b^2 + ... + C(k+1,k-1) * a^2 * b^(k-1) + C(k+1,k) * a^1 * b^k + C(k+1,k+1) * a^0 * b^(k+1)
The result matches the form of the binomial theorem for n = k+1, so the equation holds true for all positive integers n.
Therefore, the binomial theorem has been proven.