write down the first four terms of the binomial expansion of (1-y)^8 in ascending power of y.By putting y=1/2x(1-x) in your expansion, find the value of p and q if [1-1/2x(1-x)]^8=1-4x+px^2+qx^3
so plug in the Binomial Theorem, and you get
(1-y)^8 = 1 - C(8,1)y + C(8,2)y^2 - C(8,3)y^3 + ...
= 1 - 8y + 28y^2 - 56y^3 + ...
So now let y = 1/2 x(1-x) and you get
(1-y)^8 = 1 - 8(1/2 x(1-x)) + 28(1/2 x(1-x))^2 - 56(1/2 x(1-x))^3 + ...
= 1 - 4x(1-x) + 7x^2(1-x)^2 - 7x^3(1-x^3) + C(8,4)/2^4 x^4(1-x)^4 - ...
Clearly, only the first four terms will involve powers of x up to 3, so that gives
= 1 - 4x + 11x^2 - 21x^3 + ...
please solve for me
To find the first four terms of the binomial expansion of (1-y)^8, we can use the Binomial Theorem. The Binomial Theorem states that for any positive integer n:
(1 + x)^n = C(n,0) * x^0 + C(n,1) * x^1 + C(n,2) * x^2 + ... + C(n,n) * x^n
Where C(n,k) represents the number of ways to choose k items from a set of n distinct items and is given by the formula:
C(n,k) = n! / (k! * (n-k)!)
In this case, we have (1-y)^8, so we substitute y for x, and n = 8:
(1 - y)^8 = C(8,0) * y^0 + C(8,1) * y^1 + C(8,2) * y^2 + C(8,3) * y^3 + ...
Let's calculate the first four terms:
C(8,0) * y^0 = 1 * 1^0 = 1
C(8,1) * y^1 = 8 * 1^1 = 8
C(8,2) * y^2 = 28 * 1^2 = 28
C(8,3) * y^3 = 56 * 1^3 = 56
Therefore, the first four terms of the binomial expansion of (1-y)^8 are:
1 - 8y + 28y^2 - 56y^3
Now, let's substitute y = 1/2x(1-x) into the expansion:
1 - 8(1/2x(1-x)) + 28(1/2x(1-x))^2 - 56(1/2x(1-x))^3
Simplifying further,
1 - 4(1-x)/x + 7(1-x)^2/x^2 - 14(1-x)^3/x^3
Expand and collect like terms:
1 - 4/x + 4/x - 7/x + 7x/x^2 - 14x^2/x^3 + 14x^3/x^3
Simplifying,
1 - 7/x + 7x/x^2 - 14x^2/x^3 + 14x^3/x^3
1 - (7/x) + (7x/x^2) - (14x^2/x^3) + 14
Comparing this expression with the given expression 1 - 4x + px^2 + qx^3, we can identify the values of p and q:
p = 7
q = -14
Therefore, p = 7 and q = -14.
To find the first four terms of the binomial expansion of (1-y)^8 in ascending power of y, we can use the binomial theorem.
The binomial theorem states that for any real number n and any real number x, the expansion of (a + b)^n can be expressed as:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ...
Where C(n, k) represents the binomial coefficient, given by C(n, k) = n! / (k!(n-k)!).
In our case, (1-y)^8 can be expressed as:
(1-y)^8 = C(8, 0) * 1^8 * y^0 + C(8, 1) * 1^7 * y^1 + C(8, 2) * 1^6 * y^2 + C(8, 3) * 1^5 * y^3 + ...
Calculating the values:
C(8, 0) = 1
C(8, 1) = 8
C(8, 2) = 28
C(8, 3) = 56
Thus, the first four terms of the binomial expansion of (1-y)^8 in ascending power of y are:
1 - 8y + 28y^2 - 56y^3
Now, to find the value of p and q in the given expression [1 - 1/2x(1-x)]^8 = 1 - 4x + px^2 + qx^3, we can substitute y = 1/2x(1-x) into the expansion.
Replacing y with 1/2x(1-x) in the expansion (1-y)^8, we have:
[1 - 1/2x(1-x)]^8 = 1 - 8 * (1/2x(1-x)) + 28 * (1/2x(1-x))^2 - 56 * (1/2x(1-x))^3
To simplify this expression, we can start by simplifying the terms:
1/2x(1-x) = (1 - x) / (2x)
(1/2x(1-x))^2 = [(1 - x) / (2x)]^2 = (1 - x)^2 / (4x^2)
(1/2x(1-x))^3 = [(1 - x) / (2x)]^3 = (1 - x)^3 / (8x^3)
Now, substituting these values into the expansion, we have:
[1 - 1/2x(1-x)]^8 = 1 - 8 * (1 - x) / (2x) + 28 * (1 - x)^2 / (4x^2) - 56 * (1 - x)^3 / (8x^3)
Simplifying further:
= 1 - 4(1 - x) / x + 7(1 - x)^2 / (2x^2) - 7(1 - x)^3 / (2x^3)
Thus, the value of p is 7/2 and the value of q is -7/2.