# Finding the value of combination using integration

1. Jul 29, 2013

### songoku

1. The problem statement, all variables and given/known data
By using integration, find the value of
$$100C0+\frac{1}{2}100C1+\frac{1}{3}100C2+\frac{1}{4}100C3+...+\frac{1}{101}100C100$$

2. Relevant equations
Integration
Combination

3. The attempt at a solution
I don't even have any ideas to start. Can this really be solved by using integration?

Thanks

2. Jul 29, 2013

### haruspex

The trick is to turn the sum into a power series, then perform a calculus operation on it which gets rid of those nasty 1/r factors.

3. Jul 29, 2013

### lurflurf

recall that for k=1,2,...

$$\frac{1}{k}=\int_0^1 \! x^{k-1} \, \mathrm{dx} \\ \text{or} \\ \frac{1}{k}=\int_0^\infty \! e^{-k \, x} \, \mathrm{dx}$$

use binomial theorem

4. Jul 29, 2013

### songoku

I don't think I get both of your hints

The question can be written as:

$$\Sigma_{r=0}^{100}~100Cr . \frac{1}{r+1}$$

Then, by using Maclaurin series:
$$\frac{1}{r+1}=1-r+r^2-r^3+...=\Sigma_{i=0}^{∞}(-r)^i$$

I am not sure what I am doing and the direction I am heading to....

Thanks

5. Jul 29, 2013

### haruspex

My hint was to introduce a variable, something like (but not exactly):
$$\Sigma_{r=0}^{100}~100Cr . \frac{1}{r+1} s^r$$

6. Jul 30, 2013

### ehild

Use binomial theorem to write up 100 (1+x)100. Then integrate it, from 0 to 1. What do you get?

ehild

7. Jul 30, 2013

### lurflurf

my hint was

$$\sum_{k=0}^{100} {100 \choose k} \, \frac{1}{k+1}=\sum_{k=0}^{100} {100 \choose k} \, \int_0^1 \! x^k \, \mathrm{dx}=\int_0^1\sum_{k=0}^{100} {100 \choose k} x^k \, 1^{100-k} \, \mathrm{dx}$$
use the binomial theorem

8. Aug 14, 2013

### songoku

I am really sorry for taking a long time to reply

Sorry I still don't get the hint

I don't get the last part. Where 1100-k comes from? And also how to use the binomial theorem for your equation? I only know that binomial theorem is used for expanding. In your equation, what is the term that can be expanded using binomial theorem?

Wow, expanding (1 + x)100 then integrating it from 0 to 1 gives me the same result as calculating

$$\sum_{k=0}^{100} {100 \choose k} \, \frac{1}{k+1}$$

So, I can write

$$\sum_{k=0}^{100} {100 \choose k} \, \frac{1}{k+1}=\int_{0}^{1}{(1+x)^{100}}dx$$

How can we know such sigma form can be written in simple integral form? Do I have to memorize certain form or expression?

Thanks

9. Aug 15, 2013

### Dick

Just use the binomial theorem on (1+x)^100 and integrate terms. Knowing the binomial theorem is all you have to memorize. Knowing how it relates to to your problem is more of a hunch thing.

10. Aug 15, 2013

### lurflurf

I am using the binomial theorem backwards to combine the terms. I insert 1^(100-k) to make the equation exactly fit the binomial theorem. Of course 1^(100-k)=1 so it does not change anything.
$$\sum_{k=0}^{100} {100 \choose k} x^k \, 1^{100-k}=(1+x)^{100}$$

11. Aug 16, 2013

### songoku

Oh ok. Now I can see clearer the connection between stating the question in sigma form and then introducing integration to solve the problem.

Thanks a lot for all the help I got in this thread