Binomial Theorem

[skyd92]

I am having a really hard time not only understanding the arithmetic of the problem, but what the problem is asking for. I don't know where to begin at all.

Problem: "Find the coefficient of the indicated term in the expansion of the given power of a binomial. Identify the missing exponents in the term that contains: ((x^?)y^9); (x + y)^11"

I need some serious help.

 

[BigGlenntheHeavy]

\(\displaystyle (x+y)^{11} \ = \ \sum_{k=0}^{11}\big(\combination_{11}^{k}\big)x^{11-k}y^{k}\)



Well, when k = 9, we have\(\displaystyle \big(\combination_{11}^{9}\big)x^{2}y^{9} \ = \ 55x^{2}y^{9}.\)

 

[skyd92]

Thank you very much. I am just learning this concept and I thank you for helping me out.

 

[Loren]

I am having a really hard time not only understanding the arithmetic of the problem, but what the problem is asking for. I don't know where to begin at all.

Problem: "Find the coefficient of the indicated term in the expansion of the given power of a binomial. Identify the missing exponents in the term that contains: ((x^?)y^9); (x + y)^11"

I need some serious help.

http://upload.wikimedia.org/math/5/6/a/ ... de2262.png

If you take a look at the expansions at this site, you will notice that the exponent of the y term is always one less than the number of the term in which it appears. y[sup:wgc24clj]5[/sup:wgc24clj] is in the 6th term, y[sup:wgc24clj]3[/sup:wgc24clj] is in th fourth term, etc. Likewise you can see a pattern for the exponent of the x factors. In the first term the exponent of the x factor is the same as the exponent of the binomial and each subsequent term produces and exponent that is 1 less than the preceding term.
As an example, if you start with (x + y)[sup:wgc24clj]7[/sup:wgc24clj], the fifth term will have the factors x[sup:wgc24clj]3[/sup:wgc24clj]y[sup:wgc24clj]4[/sup:wgc24clj].
Hope this helps.

 

[skyd92]

thanks! I did find that very helpful.