How Does Differentiating the Binomial Theorem Help Solve Summations?

[Punkyc7]

1* 100C1 + 2* 100C2 ... + 100 * 100C100=100(2^{99})So I was thinking I would use a symmetric argument and I got

100 *( .5 \Sigma (\stackrel{100}{k}))

where k is 1 and the top of the sigma is 100... How do you make the summation in Latex? which would get me the right answer if I used a zero instead of the one from the Binomial expansion formula. If you start at zero though wouldn't I be including an extra 1 from 100C0. My problem is do you have to start at zero and if not how can you change it to start at any arbitrary number without changing the the sum.

 

[LCKurtz]

Use the quote button to see how I have written your problem:

\sum_{k=1}^{100} k\binom{100}{k} = 100(2^{99})

Here's a hint: Start with the binomial expansion of (1+x)¹⁰⁰ and differentiate it. See if that gives you any ideas.

 

[Punkyc7]

There's nothing wrong with your original idea of reversing the sum and adding. 0*100C0 doesn't contribute anything. You may as well include it and start the sum at 0.

Ok so It does work I was just worried about the extra one that might get thrown in.


For the differentiating one I got

\sum_{k=0}^{100} k\binom{100}{k} (100-k) x^{99-k}


but that doesn't look anything like 100(2^99)

 

[LCKurtz]

Ok But was there anything that I did that was wrong above?

I can't make any sense of what you wrote. For one thing it isn't an equation since there is no = sign, and I don't know what your "symmetric argument" is.

Is there any clever way today that with about writing all the terms out because I am not sure how to differentiate 100 terms

Write the binomial expansion with a summation. You can differentiate under the summation.

 

[Dick]

1* 100C1 + 2* 100C2 ... + 100 * 100C100=100(2^{99})


So I was thinking I would use a symmetric argument and I got

100 *( .5 \Sigma (\stackrel{100}{k}))

where k is 1 and the top of the sigma is 100... How do you make the summation in Latex?


which would get me the right answer if I used a zero instead of the one from the Binomial expansion formula. If you start at zero though wouldn't I be including an extra 1 from 100C0. My problem is do you have to start at zero and if not how can you change it to start at any arbitrary number without changing the the sum.

There's nothing wrong with your original idea of reversing the sum and adding. 0*100C0 doesn't contribute anything. You may as well include it and start the sum at 0.

 

[Punkyc7]

There's nothing wrong with your original idea of reversing the sum and adding. 0*100C0 doesn't contribute anything. You may as well include it and start the sum at 0.

Ok so It does work I was just worried about the extra one that might get thrown in.For the differentiating one I got

\sum_{k=0}^{100}\binom{100}{k} (100-k) x^{99-k}[/itex]but that doesn't look anything like 100(2^99), How do I know what x is?

 

[LCKurtz]

You have to have an equals sign and something on each side to have an equation. It will be easier if you think of the expansion of (x + 1)¹⁰⁰ so start with

(x + 1)^{100} = \sum_{k=0}^{100} \binom {100}{k}x^k

 

[Punkyc7]

Ok but what is the value of x... I am not sure how you came up with that

 

[Punkyc7]

You have to have an equals sign and something on each side to have an equation. It will be easier if you think of the expansion of (x + 1)¹⁰⁰ so start with

(x + 1)^{100} = \sum_{k=0}^{100} \binom {100}{k}x^k

When you differentiate the right side you get dy/dx (x + 1)^{100} = \sum_{k=0}^{100} \binom {100}{k}(k)x^(k-1)

 

[Dick]

Why didn't you differentiate both sides? And then put x=1?

 

[Punkyc7]

I didn't know x=1 that's was never given. How do you know to set it to one?

 

[Dick]

I didn't know x=1 that's was never given

The equation is true whatever x is. Why would x=1 be a good choice to put in?

 

[Dick]

I didn't know x=1 that's was never given. How do you know to set it to one?

If you put x=1 then the right side is what you want to sum, isn't it?

 

[Punkyc7]

I don't know what the 100 terms are when they are added together maybe.

So the derivative is 100(1+x)^99 and if x = 1 then it is 100(2)^99. What is the point of differentiating the right side if the left side gets our answer?

It still seems like magic that you just picked (1+x)^100

 

[Dick]

I don't know what the 100 terms are when they are added together maybe.

So the derivative is 100(1+x)^99 and if x = 1 then it is 100(2)^99. What is the point of differentiating the right side if the left side gets our answer?

It still seems like magic that you just picked (1+x)^100

You can't start out with an equation and then just differentiate one side and expect it to still be an equation, can you?? I really don't know what you are going on about. It's not magic it just worked. It's just knowing the binomial expansion. And being mildly clever about how to use it.

 

[Punkyc7]

Yes but how did you know know that (1+x)^100 would work. It just seems like out of no where that (1+x)^100 was chosen by LCKurtz. Where is the motivation that would lead you to come up with that clever idea. Aren't there many other options that when picked wouldn't work even though they look like they might?

 

[Dick]

Yes but how did you know know that (1+x)^100 would work. It just seems like out of no where that (1+x)^100 was chosen by LCKurtz. Where is the motivation that would lead you to come up with that clever idea. Aren't there many other options that when picked wouldn't work even though they look like they might?

You seem to know 0C100+1C100+...+100C100=2^100. How did you prove that?

 

[LCKurtz]

Use the quote button to see how I have written your problem:

\sum_{k=1}^{100} k\binom{100}{k} = 100(2^{99})

Here's a hint: Start with the binomial expansion of (1+x)¹⁰⁰ and differentiate it. See if that gives you any ideas.

Yes but how did you know know that (1+x)^100 would work. It just seems like out of no where that (1+x)^100 was chosen by LCKurtz. Where is the motivation that would lead you to come up with that clever idea. Aren't there many other options that when picked wouldn't work even though they look like they might?

The motivation comes from your original problem which, although you never stated it explicitly, was to prove the above statement. If you didn't have the k in there that looks like a binomial expansion of (1+x) with x = 1. That's where you would get the 2 and why there are no terms like x^ky^n-k in the expansion. And the way to get k out in front looks like a derivative of x^k. You just need familiarity with the binomial expansion and derivatives to see that.