Proving Pascal's Formula for Combinations: Using Pascal's Triangle and Formula

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The discussion focuses on proving Pascal's formula for combinations, specifically the equation that sums combinations from k to n. Participants clarify the notation and emphasize the need to use Pascal's triangle and mathematical induction for the proof. The proof involves three steps: establishing a base case, assuming the formula holds for a specific n, and proving it for n + 1. There is some confusion about the problem's wording, but the main approach remains centered on induction. Overall, the conversation highlights the importance of understanding the combinatorial identities involved in the proof.
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Prove "Pascals formula"

Homework Statement


Show that (k)+(k+1)+...+(n) = (n+1) it's supposed to be combinations
(k) (k ) (k) (k+1)

That's the question in my math book. I guess you need to use Pascals triangle and Pascals formula but unfortionately I don't know how to implement them here.
 
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please define what are these notations
what is the meaning of (k) ?
 
lalbatros, he means show that:

{{k}\choose{k}} + {{k+1}\choose{k}} + \dots + {{n}\choose{k}} = {{n+1}\choose{k+1}}

The second line of (k)'s and the (k+1) is supposed to line up underneath the stuff on the previous line. I'm about to hit the space bar 10 times. I just hit it 10 times, I see a big gap between "times." and "I just" but when I post this message, you'll only see one space.

Elruso, use Pascal's triangle on the right side of your equation together with induction.
 
Well, n=1,2,3,4... and k=1,2,3,4...
 
Elruso said:
Well, n=1,2,3,4... and k=1,2,3,4...
I don't know what your point is. Do induction on n. The base case is the trivial fact that {{k}\choose{k}} = {{k+1}\choose{k+1}}.
 
Elruso said:

Homework Statement


Show that (k)+(k+1)+...+(n) = (n+1) it's supposed to be combinations
(k) (k ) (k) (k+1)

That's the question in my math book. I guess you need to use Pascals triangle and Pascals formula but unfortionately I don't know how to implement them here.

Are you sure the problem is copied correctly? I think I should have read:
\left( \begin{array}{c} k \\ k \end{array} \right) + \left( \begin{array}{c} k + 1 \\ k \end{array} \right) + \left( \begin{array}{c} k + 2 \\ k \end{array} \right) + ... + \left( \begin{array}{c} k + n \\ k \end{array} \right) = \left( \begin{array}{c} k + n + 1 \\ k + 1 \end{array} \right)

As AKG's pointed out, you need a Proof By Induction. Proof By Induction requires 3 steps, i.e:
Step 1: Start with n = 0, or 1, or 2, or whatever according to what the problem asks you to do (in this case, you should choose n = 0), to see if it's correct.
Step 2: Assume the statement is true for k = n.
Step 3: prove that if it holds true for k = n, then it would also hold true for k = n + 1.

Can you go from her? :)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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