Write the term of the binomial series. Simplify.
a. The term with b^5 in (a-b)^9
b. The seventh term of (x^1/2 + x^-1/2)^8
C(9,6)a^4(-b)^5 = -84a^4b^5
C(8,6)(x^1/2)^2*x^(-1/2)^6 = 28x^1*x^-3 = 28/x^2
C(9,5)a^4(-b)^5 = -126a^4b^5
To find the term with b^5 in the expansion of (a-b)^9, we need to use the binomial theorem.
The binomial theorem states that the general term of the expansion of (a-b)^n is given by:
C(n, k) * a^(n-k) * (-b)^k,
where C(n, k) is the binomial coefficient or "n choose k" and is given by:
C(n, k) = n! / (k!(n-k)!).
Now, let's find the term with b^5 in (a-b)^9.
a.
Using the formula, the general term of the expansion of (a-b)^9 is:
C(9, k) * a^(9-k) * (-b)^k.
To find the term with b^5, we need the exponent of b to be 5. So, we set k = 5.
The term with b^5 is then:
C(9, 5) * a^(9-5) * (-b)^5.
Simplifying further, we have:
C(9, 5) * a^4 * (-b)^5.
Now, let's simplify the binomial coefficient:
C(9, 5) = 9! / (5!(9-5)!).
C(9, 5) = (9 * 8 * 7 * 6 * 5!) / (5! * 4!).
The factorials cancel out, leaving:
C(9, 5) = 9 * 8 * 7 * 6 / 4!.
Finally, the term with b^5 is:
(9 * 8 * 7 * 6 / 4!) * a^4 * (-b)^5.
b.
To find the seventh term of (x^(1/2) + x^(-1/2))^8, we will again use the binomial theorem.
The general term of the expansion of (x^a + x^b)^n is given by:
C(n, k) * (x^a)^(n-k) * (x^b)^k.
Now, let's find the seventh term.
In (x^(1/2) + x^(-1/2))^8, we have a = 1/2 and b = -1/2.
Using the formula, the seventh term will have k = 6 since the counting starts from 0.
The seventh term is then:
C(8, 6) * (x^(1/2))^(8-6) * (x^(-1/2))^6.
Simplifying, we have:
C(8, 6) * x^(8-6) * x^(-3).
Now, let's simplify the binomial coefficient:
C(8, 6) = 8! / (6!(8-6)!).
C(8, 6) = (8 * 7 * 6!) / (6! * 2!).
The factorials cancel out, leaving:
C(8, 6) = 8 * 7.
Finally, the seventh term is:
(8 * 7) * x^2 * x^(-3).
Simplifying further, we have:
56 * x^(-1).
Therefore, the seventh term of (x^(1/2) + x^(-1/2))^8 is 56 / x.