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

  • Thread starter Elruso
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In summary, the problem asks for a formula that takes in an integer and produces an integer. Pascals triangle and Pascals formula are notations used in this problem, but Elruso is not sure how to use them and needs help.
  • #1
Elruso
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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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  • #2
please define what are these notations
what is the meaning of (k) ?
 
  • #3
lalbatros, he means show that:

[tex]{{k}\choose{k}} + {{k+1}\choose{k}} + \dots + {{n}\choose{k}} = {{n+1}\choose{k+1}}[/tex]

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.
 
  • #4
Well, n=1,2,3,4... and k=1,2,3,4...
 
  • #5
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 [itex]{{k}\choose{k}} = {{k+1}\choose{k+1}}[/itex].
 
  • #6
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:
[tex]\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)[/tex]

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? :)
 

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

1. What is Pascal's formula?

Pascal's formula, also known as Pascal's triangle, is a mathematical formula that shows the coefficients of the binomial expansion. It was discovered by French mathematician Blaise Pascal in the 17th century.

2. How do you prove Pascal's formula?

Pascal's formula can be proven using mathematical induction. This method involves showing that the formula holds true for the first few cases (usually the base case and one or two other cases) and then proving that if the formula holds true for a particular case, it also holds true for the next case.

3. What is the significance of Pascal's formula?

Pascal's formula has many applications in mathematics, including in combinatorics, probability, and algebra. It is also used in fields such as physics, engineering, and computer science. It allows for the quick calculation of binomial coefficients and the expansion of binomials.

4. Can you provide an example of how Pascal's formula is used?

Sure, for example, if we want to expand (a+b)^4 using Pascal's formula, we would look at the fourth row of the triangle, which contains the coefficients 1, 4, 6, 4, and 1. We would then multiply these coefficients by the appropriate powers of a and b to get the expanded form: a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.

5. Are there any limitations to Pascal's formula?

Pascal's formula only works for binomial expansions, meaning expressions with two terms. It also assumes that the variables are raised to positive integer powers. Additionally, it only gives the coefficients of the expansion, not the full expanded form. Therefore, it may not be applicable to more complex expressions or equations.

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