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

(a+b)^n

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

x^4 y^3

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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