Posted by Shreya on Friday, January 18, 2013 at 10:00am.
andrew is working on a binomial problem (1/2x-y)^7. He asks u to :
i) what the coefficient of the 6th term
ii) find the term that has the variabes x^47^3 (ex. 34x^4y^3)
iii) find the first three terms of the expansion
I really need help with that ^ . I know the powers of A decrease with the terms, and the powers of b increase. So I think I get like 7P3 or something I just don't know where to go from there. The answer is supposed to be 21/4 for i) -35/10x^4y^3 and for iii) 1/128x^7-1/64x^6y+1/32x^5y^2.
- Pre Calculus 12 urgent - Steve, Friday, January 18, 2013 at 10:37am
The coefficients are
1 7 21 35 35 21 7 1
So, the 6th term is (21)(1/2 x)^2 (-y)^5 = -21/4 x^2 y^5
The 4th term has y^3
In the expansion, the n+1st term is
C(7,n) (1/2 x)^n (-y)^(7-n)
- Pre Calculus 12 urgent - Reiny, Friday, January 18, 2013 at 11:07am
In your text or notes you should see the following:
for the expansion
term(r+1) = C(n,r)(a^(r-n))(b^r)
e.g. for (3+x)^12
term(8) = C(12,7) (3^5) (x^7)
so you have ( (1/2)x - y)^7 or (x/2 - y)^7
term(6) = C(7,5) (x/2)^2 (-y)^5
= - 21 x^2 (1/2)^2 y^5
= - (21/4) x^2 y^5
making the coefficient -21/4 , you have a positive
There is no term containing x^4 7^3, you probably meant
I will assume that, since it would match the pattern
To have y^3, r=3 and it would be term(4)
term(4) = C(7,3) (x/2)^4 (-y)^3
= 35 x^4 (1/2)^4 (-y)^3
= - (35/16)x^4 y^3 , again you are wrong in the coefficient
first three terms:
C(7,0) (x/2)^7 (-y)^0 + C(7,1) (x/2)^6 (-y) + C(7,2) (x/2)^5 (-y)^2 + ...
= 1 x^7 (1/2)^7 + 7 (x/2)^6 (-y) + 21 (x/2)^5 (-y)^2 + ..
= (1/128)x^7 - (7/64)x^6 y + (21/32)x^5 y^2 + ..
I have no clue where you are getting your answers from, or what method you are using, but you will definitely have to study this topic a bit more.
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