Provide a step-by-step derivation of the formula for the nth term of the binomial expansion (a + b)^n.
user
Sure, I'd be happy to provide a step-by-step derivation of the formula for the nth term of the binomial expansion (a + b)^n.

The binomial expansion of (a + b)^n is given by:

(a + b)^n = a^n + nC1(a^(n-1)b) + nC2(a^(n-2)b^2) + ... + nCk(a^(n-k)b^k) + ... + b^n

where C means combination or binomial coefficient.

1. Let's first understand how the coefficients (nC0, nC1, nC2, ..., nCk, ..., nCn) are derived. We start by writing down the expression (a + b) and then multiply both sides by (a + b) itself n times:

(a + b) × (a + b) × ... × (a + b) = a^n + nC1 a^(n - 1)b + nC2 a^(n - 2)b^2 + ... + nCk a^(n - k)b^k + ... + nCn b^n

2. Now, multiply both sides of the above equality termwise. The LHS gives:

   a^n + b^n + (a^(n-1)b) + (a^(n-2)b^2) + ... + (a^(n-k)b^k) + ... + (ab^(n-1))

3. The terms on the RHS can be further expanded using binomial expansion:

   a^(n - 1)b = a^(n - 2)b^2 + C(n-1, 1) * a^(n-2) * b * (b - a)

   a^(n - 2)b^2 = a^(n - 3)b^3 + C(n-2, 1) * a^(n-3) * (b^2 * (b - a)) + C(n-2, 2) * a^(n-3) * (b * (b - a) * (b - 2a))

   ......(Continuing like this)......

4. Now after the above step, you will have a chain that will give only the coefficients in permutation.

5. The process of arriving at the binomial coefficients that appear in the expansion is termed "Pascal's Triangle." It's a triangular array of numbers where each term is the sum of the two immediately above it. Eg:

      1
    1   1
  1   3   1
1   4   6   4   1

6. Using Pascal's Triangle, we can calculate these coefficients.

7. So, for the nth term:

   [nC(n-r)]a^(n-r)b^r

   Where r is the value from 1 to n.

8. NCR is the binomial coefficient that tells us how many ways we can choose r items from n items.

9. But, the value of the nth term is equal to (nCr)a^(n-r)b^r.

10. So, the general term of the expansion is:

   [nC(n-r)]a^(n-r)b^r.

11. The sum of all these terms is (a + b)^n.

12. Mathematically, it is denoted by the notation (a+b)^n = a^n + nC1(a^n-1)*b + nC2(a^n-2)*(b^2) + ... + C(n, k)(a^n-k)*b^k + ... + b^n

I hope that helps. Let me know if you have any more doubts.
assistant
Provide a step-by-step derivation of the formula for the nth term of the binomial expansion (a + b)^n.
user
Sure, I'd be happy to help with that. The formula for the nth term of the binomial expansion (a + b)^n is given by the Binomial Theorem, which is a result from elementary algebra. The formula is:

T_(n+1) = ^nC_r * a^(n-r) * b^r

where:
- T_(n+1) is the (n+1)th term of the expansion,
- n is the power to which the binomial (a + b) is raised,
- r is the position of the term in the expansion (starting from 0 for the first term),
- ^nC_r (read as "n choose r") is the binomial coefficient, which is given by the formula n! / [r!(n-r)!] where "!" denotes factorial.

Here's a step-by-step derivation of the formula:

1. Start with the binomial (a + b).

2. Raise it to the nth power: (a + b)^n.

3. According to the Binomial Theorem, the expansion of this power will have (n+1) terms.

4. Each term in the expansion is of the form T_(r+1) = ^nC_r * a^(n-r) * b^r.

5. The binomial coefficient ^nC_r is the number of ways to choose r successes (b's) out of n trials (a + b's). It is given by the formula n! / [r!(n-r)!].

6. The term a^(n-r) represents the (n-r) failures (a's) in the binomial trial, and b^r represents the r successes (b's).

7. Multiplying these together gives the general term of the binomial expansion.

So, for example, the 3rd term (r=2) of the expansion of (a + b)^5 is given by:

T_3 = ^5C_2 * a^(5-2) * b^2 = (5! / [2!(5-2)!]) * a^3 * b^2 = 10a^3b^2.

This formula can be derived rigorously using principles of combinatorics and mathematical induction, but that's beyond the scope of this explanation.
assistant