Finding the value of combination using integration

songoku
Messages
2,467
Reaction score
382

Homework Statement


By using integration, find the value of
100C0+\frac{1}{2}100C1+\frac{1}{3}100C2+\frac{1}{4}100C3+...+\frac{1}{101}100C100


Homework Equations


Integration
Combination


The Attempt at a Solution


I don't even have any ideas to start. Can this really be solved by using integration?

Thanks
 
Physics news on Phys.org
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.
 
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
 
  • Like
Likes 1 person
haruspex said:
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.

lurflurf said:
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

I don't think I get both of your hints :redface:

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
 
songoku said:
I don't think I get both of your hints :redface:

The question can be written as:

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

My hint was to introduce a variable, something like (but not exactly):
\Sigma_{r=0}^{100}~100Cr . \frac{1}{r+1} s^r
 
  • Like
Likes 1 person
Use binomial theorem to write up 100 (1+x)100. Then integrate it, from 0 to 1. What do you get?


ehild
 
  • Like
Likes 1 person
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
 
I am really sorry for taking a long time to reply

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

Sorry I still don't get the hint :redface:

lurflurf said:
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

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?

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


ehild

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
 
songoku said:
I am really sorry for taking a long time to reply



Sorry I still don't get the hint :redface:



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

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.
 
  • Like
Likes 1 person
  • #10
songoku said:
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?
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.
$$\sum_{k=0}^{100} {100 \choose k} x^k \, 1^{100-k}=(1+x)^{100}$$
 
  • Like
Likes 1 person
  • #11
Dick said:
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.

lurflurf said:
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.
$$\sum_{k=0}^{100} {100 \choose k} x^k \, 1^{100-k}=(1+x)^{100}$$

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 :smile:
 
Back
Top