Find the specified term of the expansion in each of the following:
a.6th term of (3x + y)^9
b.Middle term of (3x+y^3)^8
c.The term in (x + 2y)^10 that involves x^7
looks like you are studying the binomial theorem
You MUST memorize some basic formulas
e.g.
for (a+b)^n
term(r+1) = C(n,r) (a^r)(b^(n-r))
so for (3x+9)^9
term6 = C(9,5)(3x)^5 y^4
= 126(243x^5)y^4
= 30618 x^5 y^4
b) what is the middle term for (3x + y^3)^8 ?
c) let that term be term(r+1)
= C(10,r) x^r (2y)^(10-r)
we are interested in the term containing x^7
so r = 7 and the term is
C(10,7) x^7 (2y)^3
= 120 x^7 8y^3
= 960 x^7 y^3
To find the specified term of an expansion, we can use the Binomial Theorem. The Binomial Theorem allows us to expand expressions of the form (a + b)^n, where a and b are constants, and n is a positive integer.
a) To find the 6th term of (3x + y)^9, we can use the Binomial Theorem formula:
Coefficient * (a)^n-k * (b)^k
In this case, a = 3x, b = y, and n = 9. We want to find the 6th term, which means k = 6-1 = 5 (using 1-based indexing).
Using the formula, the 6th term can be calculated as follows:
Coefficient = C(n, k) = C(9, 5) = 9! / (5! * (9-5)!) = 9! / (5! * 4!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126
Term = Coefficient * (a)^(n-k) * (b)^k
= 126 * (3x)^(9-5) * (y)^5
= 126 * (3^4 * x^4) * (y^5)
= 126 * 81 * x^4 * y^5
= 10206x^4y^5
Therefore, the 6th term of (3x + y)^9 is 10206x^4y^5.
b) To find the middle term of (3x + y^3)^8, we need to find the term with the highest degree when expanded. Since (3x) and (y^3) have different degrees, we need to distribute them in all possible combinations.
Using the Binomial Theorem, the middle term will have equal exponent for both (3x) and (y^3). In this case, the highest exponent is (y^3)^4, which equals y^12.
Using the formula, the middle term can be calculated as follows:
Coefficient = C(n, k) = C(8, 4) = 8! / (4! * (8-4)!) = 8! / (4! * 4!) = (8 * 7 * 6 * 5) / (4 * 3 * 2 * 1) = 70
Term = Coefficient * (a)^(n-k) * (b)^k
= 70 * (3x)^(8-4) * (y^3)^4
= 70 * (3^4 * x^4) * (y^12)
= 70 * 81 * x^4 * y^12
= 5670x^4y^12
Therefore, the middle term of (3x + y^3)^8 is 5670x^4y^12.
c) To find the term in (x + 2y)^10 that involves x^7, we can again use the Binomial Theorem.
In this case, a = x, b = 2y, and n = 10. We are looking for the term involving x^7, which means k = 7.
Using the formula, the term can be calculated as follows:
Coefficient = C(n, k) = C(10, 7) = 10! / (7! * (10-7)!) = 10! / (7! * 3!) = (10 * 9 * 8) / (3 * 2 * 1) = 120
Term = Coefficient * (a)^(n-k) * (b)^k
= 120 * (x)^(10-7) * (2y)^7
= 120 * (x^3) * (2^7 * y^7)
= 120 * x^3 * 128 * y^7
= 15360x^3y^7
Therefore, the term in (x + 2y)^10 that involves x^7 is 15360x^3y^7.