Write down the first four terms in the binomial series for
a). (1+3x)^(-6)
b). ∛(√(8-2x))
Just use the Binomial Theorem. I'll do one. You try the other, and come back with your work if you get stuck.
(8-2x)^(1/3)
= 8^(1/3) (-2x)^0 + (1/3) 8^(-2/3) (-2x)^1 + (1/3)(-2/3)/2 8^(-5/3) (-2x)^2 + (1/3)(-2/3)(-5/3)/3! 8^(-8/3) (-2x)^3 + ...
= 2 - x/6 - x^2/72 - 5x^3/2592 - ...
a) To find the first four terms in the binomial series for (1+3x)^(-6), we can use the binomial theorem. The general form of the binomial theorem is:
(1+x)^n = C(n,0)*1^n*x^0 + C(n,1)*1^(n-1)*x^1 + C(n,2)*1^(n-2)*x^2 + ... + C(n,r)*1^(n-r)*x^r + ...
where C(n,r) is the binomial coefficient given by C(n,r) = n! / (r!(n-r)!).
In this case, n = -6 and x = 3x. Let's write down the first four terms:
Term 1: C(-6,0)*1^(-6)*3x^0 = 1
Term 2: C(-6,1)*1^(-7)*3x^1 = -6(3x)
Term 3: C(-6,2)*1^(-8)*3x^2 = 21(9x^2)
Term 4: C(-6,3)*1^(-9)*3x^3 = -56(27x^3)
So, the first four terms in the binomial series for (1+3x)^(-6) are: 1 - 18x + 189x^2 - 1512x^3.
b) To find the first four terms in the binomial series for ∛(√(8-2x)), we can again use the binomial theorem. However, in this case, we need to simplify the expression first.
∛(√(8-2x)) can be written as ((8-2x)^0.5)^1/3.
Let's expand ((8-2x)^0.5)^1/3 using the binomial theorem:
Term 1: C(1/3,0)*((8-2x)^0.5)^(1/3)*1^0 = 1
Term 2: C(1/3,1)*((8-2x)^0.5)^(-2/3)*1^1 = (1/3)((8-2x)^0.5)^(-2/3)
Term 3: C(1/3,2)*((8-2x)^0.5)^(-5/3)*1^2 = (1/3)(-2/3)((8-2x)^0.5)^(-5/3)
Term 4: C(1/3,3)*((8-2x)^0.5)^(-8/3)*1^3 = (1/3)(-2/3)(-5/3)((8-2x)^0.5)^(-8/3)
So, the first four terms in the binomial series for ∛(√(8-2x)) are: 1 + (1/3)((8-2x)^0.5)^(-2/3) - (2/9)((8-2x)^0.5)^(-5/3) + (10/81)((8-2x)^0.5)^(-8/3).
a). To find the first four terms in the binomial series for (1+3x)^(-6), we can use the Binomial Theorem.
The Binomial Theorem states that for any real number a, b, and a positive integer n, the expansion of (a + b)^n can be written 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 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
Where C(n, k) denotes the binomial coefficient, which is given by:
C(n, k) = n! / (k! * (n - k)!)
For the given expression (1+3x)^(-6), we have a = 1, b = 3x, and n = -6.
Now, let's find the first four terms:
Term 1: The term with k = 0
C(-6, 0) * 1^(-6) * (3x)^0
= 1 * 1 * 1
= 1
Term 2: The term with k = 1
C(-6, 1) * 1^(-6) * (3x)^1
= -6 * 1 * 3x
= -18x
Term 3: The term with k = 2
C(-6, 2) * 1^(-6) * (3x)^2
= 15 * 1 * (3x)^2
= 45x^2
Term 4: The term with k = 3
C(-6, 3) * 1^(-6) * (3x)^3
= -20 * 1 * (3x)^3
= -540x^3
Therefore, the first four terms in the binomial series for (1+3x)^(-6) are: 1, -18x, 45x^2, -540x^3.
b). To find the first four terms in the binomial series for ∛(√(8-2x)), we use a similar process.
First, we can simplify the expression ∛(√(8-2x)) as (8-2x)^(1/6).
Now, we can use the Binomial Theorem to expand (8-2x)^(1/6).
Using the same approach as in part a), we can find the binomial coefficients and calculate the terms:
Term 1: The term with k = 0
C(1/6, 0) * 8^(1/6) * (-2x)^0
= 1 * 8^(1/6) * 1
= 8^(1/6)
Term 2: The term with k = 1
C(1/6, 1) * 8^(1/6 - 1) * (-2x)^1
= (1/6) * 8^(-5/6) * (-2x)
= -(1/6) * (2^(5/6)) * (2x)
= -2^(5/6) x
Term 3: The term with k = 2
C(1/6, 2) * 8^(1/6 - 2) * (-2x)^2
= (1/6 * (1/6 - 1)) * 8^(-4/6) * (-2x)^2
= (1/6 * (-5/6)) * 2^(-2/3) * (2x)^2
= (1/6)(5/6) * 2^(-2/3) * 4x^2
= (5/36) * 2^(-2/3) * 4x^2
= (10/9) * 2^(-2/3) * x^2
Term 4: The term with k = 3
C(1/6, 3) * 8^(1/6 - 3) * (-2x)^3
= (1/6 * (1/6 - 1) * (1/6 - 2)) * 8^(-3/6) * (-2x)^3
= (1/6 * (-5/6) * (-11/6)) * 2^(-1/3) * (2x)^3
= (5/36) * (11/36) * 2^(-1/3) * 8x^3
= (55/3888) * 2^(-1/3) * 8x^3
= (55/486) * 2^(-1/3) * x^3
Therefore, the first four terms in the binomial series for ∛(√(8-2x)) are:
8^(1/6), -2^(5/6) x, (10/9) * 2^(-2/3) * x^2, (55/486) * 2^(-1/3) * x^3.