组合求和级数求解：∑_{k=1}^{15} C(30,2k-1)/2k的解法及通用思路

Hey there! Let's break down how to solve this sum step by step, then talk about the general strategies you can use for similar problems.

First, your observation about $\binom{30}{29}/30 = \binom{30}{1}/30$ is spot-on—this comes from the symmetry of binomial coefficients: $\binom{n}{k} = \binom{n}{n-k}$. That symmetry is useful, but we can lean on even more powerful tools here.

Step-by-Step Solution for $\sum_{k=1}^{15} \frac{\binom{30}{2k-1}}{2k}$ #

The key trick here is converting the fraction $\frac{1}{2k}$ into an integral, since $\frac{1}{t} = \int_0^1 x^{t-1} dx$ for positive $t$. Applying this to our term:
$$\frac{1}{2k} = \int_0^1 x^{2k-1} dx$$

We can swap the order of summation and integration (valid here because all terms are positive):
$$\sum_{k=1}^{15} \frac{\binom{30}{2k-1}}{2k} = \int_0^1 \sum_{k=1}^{15} \binom{30}{2k-1} x^{2k-1} dx$$

Next, use the binomial theorem to simplify the sum inside the integral. Remember:
$$(1+x)^{30} = \sum_{m=0}^{30} \binom{30}{m} x^m$$
$$(1-x)^{30} = \sum_{m=0}^{30} \binom{30}{m} (-x)^m$$

Subtract these two equations—even-powered terms cancel out, and odd-powered terms double:
$$(1+x)^{30} - (1-x)^{30} = 2\sum_{k=1}^{15} \binom{30}{2k-1} x^{2k-1}$$

Rearrange to isolate our sum:
$$\sum_{k=1}^{15} \binom{30}{2k-1} x^{2k-1} = \frac{(1+x)^{30} - (1-x)^{30}}{2}$$

Substitute back into the integral:
$$\text{Original sum} = \int_0^1 \frac{(1+x)^{30} - (1-x)^{30}}{2} dx$$

Calculate each integral separately:

• $\int_0^1 (1+x)^{30} dx = \frac{(1+x)^{{31}}{31}\bigg|_0}1 = \frac{2^{31} - 1}{31}$
• $\int_0^1 (1-x)^{30} dx = \frac{-(1-x)^{{31}}{31}\bigg|_0}1 = \frac{1}{31}$

Combine the results:
$$\text{Original sum} = \frac{1}{2} \left( \frac{2^{31} - 1}{31} - \frac{1}{31} \right) = \frac{1}{2} \cdot \frac{2^{31} - 2}{31} = \frac{2^{30} - 1}{31}$$

Numerically, $2^{30} = 1073741824$, so $2^{30}-1 = 1073741823$, dividing by 31 gives 34636833.

General Methods for这类 Problems #

Here are the go-to strategies for sums involving binomial coefficients divided by linear terms:

• Integral Conversion
  As we used here, $\frac{1}{m} = \int_0^1 x^{m-1} dx$ turns fractions into integrals, letting you swap sum and integral to leverage the binomial theorem or generating functions. This works for any sum of the form $\sum \frac{\binom{n}{m}}{m}$.

• Binomial Theorem for Odd/Even Terms
  Using $(1+x)^n \pm (1-x)^n$ isolates sums of even or odd-indexed binomial coefficients. This is essential when dealing with sums that only include every other term (like your odd-indexed $\binom{30}{2k-1}$).

• Generating Functions
  Define a generating function where your term is a coefficient, then take derivatives or integrals to simplify the sum. For example, if you let $G(x) = \sum \frac{\binom{n}{2k-1}}{2k} x^{2k}$, then $G'(x)$ gives you the sum of $\binom{n}{2k-1}x^{2k-1}$, which we can compute with binomial theorems.

• Binomial Coefficient Symmetry
  Your initial observation about $\binom{30}{29} = \binom{30}{1}$ is part of this—$\binom{n}{k} = \binom{n}{n-k}$ can rewrite terms to make sums symmetric or easier to pair up.

内容的提问来源于stack exchange，提问作者user5722540