求和：二项式定理的改写——从形式1推导形式2的技术问询

Hey there! Let's walk through the exact algebraic steps to convert the first form of the binomial theorem to the second—no intuitive leaps required. I’ll start by clarifying the two common forms you’re referring to, then break down the derivation step by step.

First, Define the Two Forms #

Let’s assume your form (1) is the standard "ascending $b$ powers" version:
$$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n - k} b^k \tag{1}$$
And form (2) is the "ascending $a$ powers" variant:
$$(a + b)^n = \sum_{m=0}^n \binom{n}{m} a^{m} b^{n - m} \tag{2}$$

The core difference here is the index of summation and which variable’s power increases with the index. The derivation relies on variable substitution and a key property of binomial coefficients.

Step 1: Introduce a New Summation Variable #

Let’s define a new variable $m$ such that:
$$m = n - k$$
This lets us rewrite the index $k$ in terms of $m$:
$$k = n - m$$

Next, adjust the summation bounds to match the new variable:

• When $k = 0$ (the lower bound of form 1), $m = n - 0 = n$
• When $k = n$ (the upper bound of form 1), $m = n - n = 0$

Step 2: Substitute Into Form (1) #

Replace every instance of $k$ in form (1) with $n - m$:
$$(a + b)^n = \sum_{m=n}^0 \binom{n}{n - m} a^{n - (n - m)} b^{n - m}$$

Simplify each part of the summand:

1. Exponent of $a$: $n - (n - m) = m$
2. Exponent of $b$: $n - m$ (already simplified)
3. Binomial coefficient: By the symmetry property of binomial coefficients, $\binom{n}{n - m} = \binom{n}{m}$. This holds because:
   $$\binom{n}{n - m} = \frac{n!}{(n - m)! \cdot (n - (n - m))!} = \frac{n!}{(n - m)! \cdot m!} = \binom{n}{m}$$

After simplifying, our equation becomes:
$$(a + b)^n = \sum_{m=n}^0 \binom{n}{m} a^{m} b^{n - m}$$

Step 3: Reverse the Summation Order #

Addition is commutative—changing the order of terms doesn’t change the total sum. So we can reverse the bounds of summation from $m=n$ to $m=0$ to $m=0$ to $m=n$, with no change to the terms themselves:
$$(a + b)^n = \sum_{m=0}^n \binom{n}{m} a^{m} b^{n - m}$$

This is exactly form (2)!

If Your Form (2) Is a Different Variant? #

If the second form you’re referring to has a slightly different summation range (e.g., starting at $k=1$ instead of $k=0$), the same core logic applies: use variable substitution to map the original index to the new one, adjust bounds, and apply binomial coefficient properties as needed.

内容的提问来源于stack exchange，提问作者W. G.