For the expansion of (a − 3b
2
)
5
, determine the general term, and
the coefficient of a
3b
4
.
To expand (a - 3b^2)^5, we can use the Binomial Theorem. According to the Binomial Theorem, the expansion of (a - 3b^2)^n can be written as:
(a - 3b^2)^n = C(n, 0) * a^n * (-3b^2)^0 + C(n, 1) * a^(n-1) * (-3b^2)^1 + C(n, 2) * a^(n-2) * (-3b^2)^2 + ... + C(n, n-1) * a^1 * (-3b^2)^(n-1) + C(n, n) * a^0 * (-3b^2)^n
In this case, for the expansion of (a - 3b^2)^5, we have n = 5.
The general term of the expansion will have the form C(5, k) * a^(5-k) * (-3b^2)^k, where k is the power of (-3b^2).
To find the coefficient of a^3b^4, we need to find the term in the expansion where the powers of a and b in each term add up to 3 and 4 respectively.
We can set up the equation:
(5 - k) + 2k = 3
5 - k + 2k = 3
5 + k = 3
k = 3 - 5
k = -2
Since k cannot be a negative number, this means that there is no term with a^3b^4 in the expansion.
Therefore, the general term of the expansion is given by C(5, k) * a^(5-k) * (-3b^2)^k, and the coefficient of a^3b^4 is 0.
To find the general term and the coefficient of a^3b^4 in the expansion of (a - 3b^2)^5, we can use the binomial theorem.
The binomial theorem states that the expansion of (a + b)^n can be written as:
(a + b)^n = nC0 * a^n * b^0 + nC1 * a^(n-1) * b^1 + nC2 * a^(n-2) * b^2 + ... + nCn * a^0 * b^n
where nCk represents the binomial coefficient, given by the formula:
nCk = n! / (k! * (n - k)!)
In this case, we have (a - 3b^2)^5. Comparing this to the binomial theorem, we can see that a = a, b = -3b^2, and n = 5.
To find the general term of this expansion, we substitute these values into the binomial theorem formula:
General term = nCk * a^(n-k) * b^k
Substituting n = 5, a = a, b = -3b^2, and k = 4, the general term becomes:
General term = 5C4 * a^(5-4) * (-3b^2)^4
Simplifying this expression:
General term = 5C4 * a^1 * (-3)^4 * (b^2)^4
We can calculate the binomial coefficient 5C4 using the formula mentioned earlier:
5C4 = 5! / (4! * (5-4)!) = 5! / (4! * 1!) = 5
Substituting this value back into the general term expression, we get:
General term = 5 * a * (-3)^4 * (b^2)^4
Simplifying this further, we have:
General term = 5 * a * 81 * b^8
Finally, to find the coefficient of a^3b^4, we compare the powers of a and b in the general term expression. We can see that for a^3b^4 to be present, the term must have a power of a = 3 and a power of b = 4.
Comparing the general term expression, we can see that the coefficient of a^3b^4 is 0, since the general term only includes a^1 and b^8.
Therefore, the coefficient of a^3b^4 in the expansion of (a - 3b^2)^5 is 0.
To determine the general term and the coefficient of a^3b^4 in the expansion of (a − 3b^2)^5, we can use the binomial theorem. The binomial theorem states that the expansion of (a + b)^n, where n is a positive integer, 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
In this case, we have (a - 3b^2)^5, where a is the first term and -3b^2 is the second term. Therefore, we can rewrite the expression as:
(a - 3b^2)^5 = C(5, 0)a^5 (-3b^2)^0 + C(5, 1)a^4 (-3b^2)^1 + C(5, 2)a^3 (-3b^2)^2 + ... + C(5, 4)a^1 (-3b^2)^4 + C(5, 5)a^0 (-3b^2)^5
Let's evaluate each individual term:
C(5, 0)a^5 (-3b^2)^0 = 1 * a^5 * 1 = a^5
C(5, 1)a^4 (-3b^2)^1 = 5 * a^4 * (-3b^2) = -15a^4b^2
C(5, 2)a^3 (-3b^2)^2 = 10 * a^3 * (-3b^2)^2 = 90a^3b^4
Now we can identify the general term and the coefficient of a^3b^4:
The general term is C(5, 2)a^3 (-3b^2)^2 = 90a^3b^4.
The coefficient of a^3b^4 is 90.