Asked by Jon
Use the Binomial Theorem to find the fifth term in the expansion (2x+3y)^5.
5!/(5-k)^x^5-kyk
5!/(5-4)!4!^x^5-4y4
(5*4*3*2)/(4*3*2*1)
120/24
5xy^4 (answer)
You may not understand the 1st 2 lines I'm sorry just let me know how to make it better.
5!/(5-k)^x^5-kyk
5!/(5-4)!4!^x^5-4y4
(5*4*3*2)/(4*3*2*1)
120/24
5xy^4 (answer)
You may not understand the 1st 2 lines I'm sorry just let me know how to make it better.
Answers
Answered by
Reiny
how about
C(5,4)(2x)^1(3y)^4
= 5*2*3^4*x*y^4
= 810xy^4
You have totally ignored the 2 and 3 in front of the x and y terms.
C(5,4) is read 5 choose 4 and has value
5!/(4!1!)
C(5,4)(2x)^1(3y)^4
= 5*2*3^4*x*y^4
= 810xy^4
You have totally ignored the 2 and 3 in front of the x and y terms.
C(5,4) is read 5 choose 4 and has value
5!/(4!1!)
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