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# help watch

1. x^6 in expanded (1+3x)^12
2. this is (3x)^6 x (12,6) - called "12 choose 6" - or (3x)^6 x [12!/6!(12-6)!]

2nd: term you`re looking for has equal powers for each part. the bimomial coefficient of this term will be:

(8 choose 1) x^(8-r) (1/x)^r

the powers must be equal, so you equate them and solve...
3. (Original post by Hasufel)
this is (3x)^6 x (12,6) - called "12 choose 6" - or (3x)^6 x [12!/6!(12-6)!]

2nd: term you`re looking for has equal powers for each part. the bimomial coefficient of this term will be:

(8 choose 1) x^(8-r) (1/x)^r

the powers must be equal, so you equate them and solve...
You need to edit that to (8 choose r)
4. (Original post by Hasufel)
this is (3x)^6 x (12,6) - called "12 choose 6" - or (3x)^6 x [12!/6!(12-6)!]

2nd: term you`re looking for has equal powers for each part. the bimomial coefficient of this term will be:

(8 choose 1) x^(8-r) (1/x)^r

the powers must be equal, so you equate them and solve...
for 2nd, how did you get r? what does that stand for?
5. (Original post by 0utdoorz)
for 2nd, how did you get r? what does that stand for?
r is any number between 1 and 8. but in order to have the power of x (which for the rth term is 8-2r) equal 0, r=4
6. of course! 8 choose r - sorry!

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