Binomial Theorem?

Question

I know the general expansion of it and how to start writing the X and Y powers in a sequence, but how do we get the coefficients? For example

(x+y)^4

X^4 Y^0 + X^3 Y^1 + X^2 Y^2 + X^1 Y^3 + X^0 Y^4

=> --------------

My question is how do we get the coefficient?
What if the term was something (2x + 3y)^4?

Thanks

Answer 1

For starters you got it wrong:

(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

because (x+y)^n = x^n + C(n,1)x^(n-1)y + C(n,2)x^(n-2y^2) +

+.....C(n,n-1)xy^(n-1) + y^n

The coefficient for the nth term for (2x +3y)^4 would be:

C(4,n-1)*2^(5-n)*3^(n-1)

Answer 2

Make the pascal triangle:
..........1
........1 1
......1 2 1
....1 3 3 1
..1 4 6 4 1
1 5 10 10 5 1
and so on.
for (x+y)^4 take a look at the 5th row:1 4 6 4 1
coefficient of X^4 Y^0 is 1
coefficient of X^3 Y^1 is 4
coefficient of X^2 Y^2 is 6
coefficient of X^1 Y^3 is 4
coefficient of X^0 Y^4 is 1

So (X+Y)^4=X^4 + 4X^3 Y^1 + 6X^2 Y^2 + 4X^1 Y^3 + Y^4
In the same way, (2x+3Y)^4=(2x)^4 + 4(2x)^3 (3y)^1 + 6(2x)^2 (3y)^2 + 4(2x)^1 (3y)^3 + (3y)^4=16x + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4
Hope this will help u.

Answer 3

The (r+1)th term will be (4Cr)[(2x)^(4-r)][(3y)^r], where 4 is the power of the Binomial expansion.

Example, if u want the 3rd term, hence 3=r+1, therefore r=2.
Then the 3rd term
= (4C2) [(2x)^2] [(3y)^2]
= {(4*3*2*1)/[(2*1)(2*1)]} (4 x^2) (9 y^2)
= (6*4*9) (x^2) (y^2)
= 216 x^2 y^2

Hence the coefficient of the 3rd term is 216.

To understand the part on (4Cr), pls look at my source.

Answer 4

X^4 Y^0 + 4C1X^3 Y^1 + 4C2X^2 Y^2 + 4C3X^1 Y^3 + 4C4X^0 Y^4

mCn = m!/(m!(m-n)!)

For (2x + 3y)^4 substitute 2x for X and 3y for Y

(2x)^4 (3y)^0 + 4C1(2x)^3 (3y)^1 + 4C2(2x)^2 (3y)^2 + 4C3(2x)^1 (3y)^3 + 4C4(2x)^0(3y)^4

Answer 5

The "Combinatorial coefficients" are referred to as the "Binomial coefficients" and are actually and lower back denoted via "B" quite of "C". the reason at the back of 2 diverse names is by technique of the fact the comparable numbers arise in 2 diverse circumstances (in actuality, greater desirable than 2). First, as you assert, they arise interior the binomial theorem. 2nd, they arise while calculating what share approaches there are to integrate a undeniable variety of goods taken out of a typical pool of n products. for example, if a usual deck of taking part in cards is dealt out to 4 human beings, one and all getting 13 taking part in cards, what share diverse hands are accessible? Our common pool, the deck, has fifty two products. To get a hand of taking part in cards, we pick 13 out of the fifty two. The variety of diverse a thank you to do it incredibly is denoted via C(fifty two,13) (examine "fifty two pick 13"). this is the comparable by way of fact the coefficient of x^13 interior the binomial develop of (a million+x)^fifty two. be conscious: in a mixture, like a hand of taking part in cards, the order of the taking part in cards makes no distinction. (it would be referred to as a "Permutation" quite of "combination" if we necessary order to be suitable).

Answer 6

You can look up binomial coefficients in a table or calculate them. See http://en.wikipedia.org/wiki/Binomial_coefficients