find the third term in the expansion of (3x-y)^6
That would of course be
6C2 (3x)^4 (-y)^2 = 15(81x^4)(y^2) = 1215x^4y^2
Oh, boy! Time for some expansion magic! Let's dive into it.
To find the third term in the expansion of (3x - y)^6, we use the Binomial Theorem. According to this prestigious theorem (it has a trophy and everything), the term with index r in the expansion of (a + b)^n is given by:
Term_r = nCr * a^(n-r) * b^r,
where nCr is the binomial coefficient, and it represents the number of ways to choose r items from a set of n items.
In this case, our original expression is (3x - y)^6. We want to find the third term, which corresponds to r = 2 (remember, we start counting with 0).
So, applying the Binomial Theorem, we have:
Term_2 = 6C2 * (3x)^(6-2) * (-y)^2,
6C2 = 6! / (2! * (6-2)!),
= 6! / (2! * 4!),
= (6 * 5 * 4!) / (2! * 4!),
= (6 * 5) / (2!),
= 15.
Combining all the powers and coefficients, we get:
Term_2 = 15 * (3x)^4 * y^2,
= 15 * 3^4 * x^4 * y^2,
= 15 * 81 * x^4 * y^2,
= 1215 * x^4 * y^2.
So, the third term in the expansion of (3x - y)^6 is 1215x^4y^2. Ta-da!
To find the third term in the expansion of (3x-y)^6, we can use the binomial theorem.
The binomial theorem states that for any binomial expression (a + b)^n, the general term in the expansion is given by:
T(r+1) = C(n, r) * a^(n-r) * b^r
Where:
T(r+1) is the (r+1)-th term,
C(n, r) is the binomial coefficient, given by C(n, r) = n! / (r! * (n - r)!),
a is the first term of the binomial expression,
b is the second term of the binomial expression,
n is the power to which the binomial expression is raised,
and r is the index of the term in the expansion (starting from r = 0 for the first term).
In this case, we have (3x-y)^6. Therefore, a = 3x, b = -y, and n = 6.
To find the third term, we need to find the term with r = 2. Plugging these values into the formula, we have:
T(2+1) = C(6, 2) * (3x)^(6-2) * (-y)^2
C(6, 2) = 6! / (2! * (6 - 2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
(3x)^(6-2) = (3x)^4 = 81x^4
(-y)^2 = (-y)^2 = y^2
Therefore, the third term in the expansion of (3x-y)^6 is:
T(3) = C(6, 2) * (3x)^(6-2) * (-y)^2 = 15 * 81x^4 * y^2 = 1215x^4y^2
To find the third term in the expansion of (3x - y)^6, we can use the Binomial Theorem. The Binomial Theorem states that the terms in the expansion of (a + b)^n can be written as:
C(n, k) * a^(n-k) * b^k
where C(n, k) is the binomial coefficient, n is the power, k is the term number (starting from 0), a and b are the variables.
In this case, our expression is (3x - y)^6, so n = 6, a = 3x, and b = -y. We want to find the third term, so k = 2.
Let's calculate the third term using the formula:
C(6, 2) * (3x)^(6-2) * (-y)^2
The binomial coefficient C(6, 2) can be calculated using the formula:
C(n, k) = n! / (k! * (n-k)!)
Now we can substitute the values:
C(6, 2) = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5 * 4 * 3 * 2 * 1) / ((2 * 1) * (4 * 3 * 2 * 1)) = 15
(3x)^(6-2) = (3x)^4 = 81x^4
(-y)^2 = y^2
Now let's substitute the values back into the formula:
C(6, 2) * (3x)^(6-2) * (-y)^2 = 15 * 81x^4 * y^2
Therefore, the third term in the expansion of (3x - y)^6 is 1215x^4y^2.