Binomial theorem?

Question

We write Binomial expansion of (1+x)^n as:
(1+x)^n = C0+C1x+C2 x^2+...........+Crx^r......+Cnx^n.
Where sign "C" means "combination".My question is Why Combination sign is used or what is the importance of Combination in Binomial theorem? Please tell me.

Answer 1

You're using the binomial theorem to find the coefficients of the various powers of x in (a+x)^n.

(a+x)^n=(a+x)(a+x)(a+x)...(a+x)(a+x)

(n parentheses). Now how are you to multiply all of this out? If you try, you'll notice that every product has n parts to it. For example, if there are 5 a's, then there have to be n-5 x's, if there are n x's, there are no a's, etc.

You can put the terms in order of how many x's are in each: none, one, two,...,n-1,n. There will be repeats. How many repeats there are for each kind of term is the coefficient for that kind of term.

How do you get each kind of term? Say you want the one with three x's. These x's could come from the first three ()'s, or from the first two and the second-to-last, or... Any such choice of three ()'s will give a term with three x's in it, and all such terms are the same, so to find the coefficient it would suffice to count how many ways you can pick 3 out of n ()'s. This number is nC3. Thinking like this is where the C-symbols come in to this problem.

Answer 2

The "Combinatorial coefficients" are also called the "Binomial coefficients" and are sometimes denoted by "B" instead of "C". The reason for two different names is because the same numbers come up in two different circumstances (in fact, more than two). First, as you say, they come up in the binomial theorem. Second, they come up when calculating how many ways there are to combine a certain number of objects taken out of a general pool of n objects. For example, if a standard deck of cards is dealt out to 4 people, each one getting 13 cards, how many different hands are possible? Our general pool, the deck, has 52 items. To get a hand of cards, we choose 13 out of the 52. The number of different ways to do this is denoted by C(52,13) (read "52 choose 13"). It is the same as the coefficient of x^13 in the binomial expansion of (1+x)^52. NOTE: in a combination, like a hand of cards, the order of the cards makes no difference. (It would be called a "Permutation" instead of "combination" if we wanted order to be relevant).

Answer 3

You should learn the basics of permutations and combinations and then come back to the binomial theorem.

Here is an excellent source that is simple and will help you understand:

http://www.themathpage.com/aPreCalc/permutations-combinations.htm

Answer 4

In your expansion the terms C0,C1,C2 ....etc. are actually nC0,nC1,nC2...etc.
where n is superscript and 0,1,2...are subscript of C.

nC0=1, nC1=n, nC2=n(n-1)/2!...etc
this follows from the general formula:

nCr=( n!)/[(r!){(n-r)!}]

note: n! stands for factorial n.

In simple (r+1)th term of a binomial expansion (x+y)^n =(nCr) (x^r) {y^(n-r)}

The combination terms are called binomial coefficients.

Answer 5

(1+x)^1=1C0+1C1x
(1+x)^2=1^2+2*1*x+x^2 =2C0+2C1x+2C2x^2
(1+x)^3=1^3+3*1^2*x+3*1*x^2+x^3
=3C0+3C1x+3C2x^2+3C3x^3
(1+x)^4=1^4+4*1^3x+6*1^2x^2+6*1x^3+x^4
=>4C0+4C1x+4C2x^2+4C3x^3+4C4x^4
thus we find that the coefficients of x can be
written using the Combination notation