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.
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)
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.
To solve these binomial problems, we can use the binomial theorem. The binomial theorem gives us a way to expand expressions of the form (a + b)^n, where a and b are variables, and n is a positive integer.
i) To find the coefficient of the 6th term, we need to know the formula for the general term of the binomial expansion. The general term is given by:
C(n, k) * a^(n - k) * b^k
Where C(n, k) is the binomial coefficient, given by C(n, k) = n! / (k! * (n - k)!). Here, n is the exponent of the binomial, and k represents the term number.
In this case, the binomial is (1/2x - y)^7, and we want to find the coefficient of the 6th term. Since the exponent is 7, we have n = 7. The kth term of the binomial expansion corresponds to the (k + 1)th term in our counting. So, for the 6th term, k = 5.
Now we can substitute these values into the formula to find the coefficient:
C(7, 5) * (1/2x)^(7 - 5) * (-y)^5
C(7, 5) = 7! / (5! * (7 - 5)!) = 7! / (5! * 2!) = 7 * 6 / 2 = 21
(1/2x)^(7 - 5) = 1/(2x)^2 = 1/(4x^2)
(-y)^5 = (-1)^5 * y^5 = -y^5
Putting it all together, the coefficient of the 6th term is 21 * 1/(4x^2) * (-y)^5, which simplifies to 21/4.
ii) To find the term that has the variables x^4 and y^3 (ex. 34x^4y^3), we need to determine the values of n, k, and the binomial coefficient C(n, k) that will result in the desired term.
In this case, we want to find the term with x^4 and y^3. So the exponents in our term should satisfy:
(1/2x)^(n - k) = (1/2x)^4 = 1/(16x^4)
(-y)^k = (-y)^3 = -y^3
Now we can solve the equations to find the values of n and k. Rearranging the first equation:
1/(2x^n) = 1/(16x^4)
2x^4 = 16x^n
x^(n - 4) = 8
Since the base is the same, the exponents must be equal. Therefore, n - 4 = 3, which gives us n = 7.
For k, we already know that k = 3.
Now we can substitute these values into the general term formula to find the term:
C(7, 3) * (1/2x)^(7 - 3) * (-y)^3
C(7, 3) = 7! / (3! * (7 - 3)!) = 7 * 6 * 5 / (3 * 2) = 35
(1/2x)^(7 - 3) = 1/(2x)^4 = 1/(16x^4)
(-y)^3 = (-1)^3 * y^3 = -y^3
Putting it all together, the term with x^4 and y^3 is 35 * 1/(16x^4) * (-y)^3, which simplifies to -35/10x^4y^3.
iii) To find the first three terms of the expansion, we need to use the general term formula.
The general term formula is:
C(n, k) * a^(n - k) * b^k
For the binomial (1/2x - y)^7, we have n = 7.
1st term (k = 0):
C(7, 0) * (1/2x)^(7 - 0) * (-y)^0 = 1 * 1/(2x)^7 * 1 = 1/(128x^7)
2nd term (k = 1):
C(7, 1) * (1/2x)^(7 - 1) * (-y)^1 = 7 * 1/(2x)^6 * (-y) = -1/(64x^6y)
3rd term (k = 2):
C(7, 2) * (1/2x)^(7 - 2) * (-y)^2 = 21 * 1/(2x)^5 * (y^2) = 1/(32x^5y^2)
So, the first three terms of the expansion are 1/(128x^7), -1/(64x^6y), and 1/(32x^5y^2).