use the binomial theorem to expand the binomial (s+2v)^5
The Binomial Theorem Quick Check:
1. A. or the one that has "-1,458d + 729" at the end of it.
2. B. or the one that has "+243v^5" at the end of it.
3. D. or "-64b^3"
100% 3/3
To expand the binomial (s + 2v)^5 using the binomial theorem, we can use the following formula:
(n choose k) * (a^(n-k)) * (b^k)
Where:
- n is the power to which the binomial is raised (in this case, n = 5)
- k is the index of the term we want to find in the expansion
- (n choose k) is the binomial coefficient, which represents the number of combinations of n things taken k at a time
- a is the variable in the first term of the binomial (in this case, a = s)
- b is the variable in the second term of the binomial (in this case, b = 2v)
Let's apply the binomial theorem to (s + 2v)^5:
Expansion of (s + 2v)^5:
First term (n choose 0) * (s^5) * (2v^0):
(5 choose 0) * (s^5) * (2v^0) = 1 * (s^5) * 1 = s^5
Second term (n choose 1) * (s^4) * (2v^1):
(5 choose 1) * (s^4) * (2v^1) = 5 * (s^4) * (2v) = 10s^4v
Third term (n choose 2) * (s^3) * (2v^2):
(5 choose 2) * (s^3) * (2v^2) = 10 * (s^3) * (4v^2) = 40s^3v^2
Fourth term (n choose 3) * (s^2) * (2v^3):
(5 choose 3) * (s^2) * (2v^3) = 10 * (s^2) * (8v^3) = 80s^2v^3
Fifth term (n choose 4) * (s^1) * (2v^4):
(5 choose 4) * (s^1) * (2v^4) = 5 * (s^1) * (16v^4) = 80sv^4
Sixth term (n choose 5) * (s^0) * (2v^5):
(5 choose 5) * (s^0) * (2v^5) = 1 * (1) * (32v^5) = 32v^5
Therefore, the expanded form of (s + 2v)^5 is:
s^5 + 10s^4v + 40s^3v^2 + 80s^2v^3 + 80sv^4 + 32v^5
To expand the binomial (s + 2v)^5 using the binomial theorem, we can use the formula:
(a + b)^n = C(n, 0) * a^(n-0) * 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, a = s, b = 2v, and n = 5.
Let's apply the formula step by step:
First, we need to calculate the binomial coefficients C(n, k) for each term in the expansion. The binomial coefficient is calculated as C(n, k) = n! / (k!(n-k)!), where n! represents the factorial of n.
For the first term, k = 0:
C(5, 0) = 5! / (0! * (5-0)!) = 1
The first term is: 1 * s^5 * (2v)^0 = s^5
For the second term, k = 1:
C(5, 1) = 5! / (1! * (5-1)!) = 5
The second term is: 5 * s^4 * (2v)^1 = 10s^4v
For the third term, k = 2:
C(5, 2) = 5! / (2! * (5-2)!) = 10
The third term is: 10 * s^3 * (2v)^2 = 40s^3v^2
For the fourth term, k = 3:
C(5, 3) = 5! / (3! * (5-3)!) = 10
The fourth term is: 10 * s^2 * (2v)^3 = 80s^2v^3
For the fifth term, k = 4:
C(5, 4) = 5! / (4! * (5-4)!) = 5
The fifth term is: 5 * s^1 * (2v)^4 = 160sv^4
For the sixth term, k = 5:
C(5, 5) = 5! / (5! * (5-5)!) = 1
The sixth term is: 1 * s^0 * (2v)^5 = 32v^5
Now we can put all the terms together:
(s + 2v)^5 = s^5 + 10s^4v + 40s^3v^2 + 80s^2v^3 + 160sv^4 + 32v^5
So, the binomial (s + 2v)^5 expands as s^5 + 10s^4v + 40s^3v^2 + 80s^2v^3 + 160sv^4 + 32v^5
The values that we need from Pascal's triangle are
1 5 10 10 5 1
(s+2v)^5
= s^5 + 5(s^4)(2v) + 10(s^3)(2v)^2 + 10(s^2)(2v)^3 + 5(s)(2v)^4 + (2v)^5
= ..
I will leave the algebraic simplification up to you