Terms

Kiran · 2007-01-03 10:49:48

write the third term of (x + z)^7 .

mathsyperson · 2007-01-03 10:59:07

That depends on which way you order your terms. The third term will either have an x² or a z², but we don't know which.

Anyway, this question just needs you to use the binomial expansion. You can find the coefficient by n!/r!(n-r)!.

Kiran · 2007-01-04 05:01:28

http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_BinomialExpansion.xml

look at the practice at the bottom of the page.
i have to do it like that?

mathsyperson · 2007-01-04 07:30:48

Yup, exactly. Although I wouldn't rely too heavily on Pascal's triangle. It's fine when you're not expanding to too high a power, but when you start getting higher than ^20 then it's very inconvenient to use.

Kiran · 2007-01-04 09:44:40

ok so look at what i did:

(x+y)^7
1st term > 1(x)^7
2nd term > 7(x)^6 . (y)^1
3rd term > 14(x)^5 . (y)^2

thats all i need right?

mathsyperson · 2007-01-05 01:55:39

You don't technically need to put the first two terms in there if you don't want to.

Also, you have the coefficient wrong. The answer should be 21x^5 y^2.

RauLiTo · 2007-01-06 07:11:32

( X + Y ) ^ 7

3rd term = 7 nCr 2 * ( X ) ^ 5 * ( y ) ^ 2 = 21 X ^ 5 y ^ 2

as mathsy said

Math Is Fun Forum

#1 2007-01-03 10:49:48

Terms

#2 2007-01-03 10:59:07

Re: Terms

#3 2007-01-04 05:01:28

Re: Terms

#4 2007-01-04 07:30:48

Re: Terms

#5 2007-01-04 09:44:40

Re: Terms

#6 2007-01-05 01:55:39

Re: Terms

#7 2007-01-06 07:11:32

Re: Terms

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