James applies the Binomial Theorem for the expansion of (2x−3)6
. Which of the following is the second term in the expansion?
−2160x4
negative 2160 x to the 4th power
−576x5
negative 576 x to the 5th power
576x5
576 x to the 5th power
2160x4
−576x5
To find the second term in the expansion of (2x-3)^6, we can use the Binomial Theorem. The Binomial Theorem states that the nth term in the expansion of (a+b)^n can be calculated using the formula:
nCr * a^(n-r) * b^r
where nCr represents the binomial coefficient, r represents the power of b in the term, and n-r represents the power of a in the term.
In this case, for the expansion of (2x-3)^6, the second term will have r = 1 (the power of b) and n-r = 6-1 = 5 (the power of a).
Using the formula, we can calculate the second term as follows:
nCr = 6C1 = 6
a^(n-r) = (2x)^5 = 32x^5
b^r = (-3)^1 = -3
The second term in the expansion is given by the product of these values:
6 * 32x^5 * -3 = -576x^5
Therefore, the second term in the expansion of (2x-3)^6 is -576x^5.
To find the second term in the expansion of (2x−3)^6 using the Binomial Theorem, we need to use the formula:
nCk * a^(n-k) * b^k
Where:
n is the exponent of the binomial (in this case, 6),
k is the term number we are looking for (in this case, 2),
a is the first term of the binomial (2x),
b is the second term of the binomial (-3),
and nCk is the binomial coefficient, which is calculated by n! / (k!(n-k)!).
Plugging in the values, we have:
n = 6,
k = 2,
a = 2x,
b = -3.
First, let's calculate the binomial coefficient:
nCk = 6C2 = 6! / (2!(6-2)!) = 6! / (2! * 4!) = (6 * 5 * 4!) / (2! * 4!) = (6 * 5) / 2 = 15.
Now, let's calculate the exponent powers:
a^(n-k) = (2x)^(6-2) = (2x)^4 = 2^4 * x^4 = 16x^4.
b^k = (-3)^2 = 9.
Finally, let's multiply all the components together:
nCk * a^(n-k) * b^k = 15 * 16x^4 * 9 = 2160x^4.
So, the second term in the expansion of (2x−3)^6 is 2160x^4.
Therefore, the correct answer is 2160x^4.