find the constant term of the expansion of (3x²+(1/x))^8
I don't know where to start =\
(1/x^8) [ 3 x^10 +1 ]^8
now find the term in x^8 in the expansion of
(3 x^10 + 1)^8
because when you multiply by (1/x^8) it will be constant.
I would use the formula for binomial coefs rather than 9 rows of Pascal
C(n,r) = n!/[r!(n-r)!]
for n = 8 and r = 2
C = 8!/[6!(2!)]
= 8*7/2 = 28
The general term in the expansion is
tr+1 = C(8,r)(3x2)8-r(1/x)r
= C(8,r) 3^(8-r) x^(16-3r)
to have a constant term the exponent of x must be zero, or
16-3r = 0
There is not integer solution for this, so the expansion does not have a "constant" term
another way to see it,...
the first term contains x^16, the second term constains x^13, .....
each term will contain an x term
To find the constant term in the expansion of the given expression, you will need to expand it using the binomial theorem. The binomial theorem states that the expansion of (a + b)^n can be written as the sum of the terms of the form (a^m)(b^(n-m)), where m takes on all values from 0 to n.
In this case, the expression is (3x^2 + (1/x))^8. Notice that we have two terms: 3x^2 and 1/x. So, we can use the binomial theorem to expand each term separately.
Let's start by expanding (3x^2)^8. According to the binomial theorem, the constant term is obtained when m = n - m in the term (3x^2)^m(1/x)^(n-m). In this case, n is 8.
To find the constant term in the expansion of (3x^2)^8, we need to set m = 4. Plugging these values into the binomial theorem, we get:
C(8, 4) * (3x^2)^4 * (1/x)^(8-4)
where C(8, 4) is the combination formula, which calculates the number of ways to choose 4 items out of 8.
Simplifying, we have:
C(8, 4) * (3^4) * (x^(2*4)) * (x^-4)
= C(8, 4) * 81 * x^8 * 1/x^4
= C(8, 4) * 81 * x^4
So, the constant term in the expansion of (3x^2)^8 is 81.
Now, let's consider the expansion of (1/x)^8. Again, using the same logic, the constant term is obtained when m = n - m in the term (3x^2)^(n-m)*(1/x)^m. In this case, n is 8.
To find the constant term in the expansion of (1/x)^8, we need to set m = 4. Plugging these values into the binomial theorem, we get:
C(8, 4) * (3x^2)^(8-4) * (1/x)^4
Simplifying, we have:
C(8, 4) * (3x^2)^4 * (1/x^4)
= C(8, 4) * 81 * x^8 * 1/x^4
= C(8, 4) * 81 * x^4 * 1/x^4
= C(8, 4) * 81
So, the constant term in the expansion of (1/x)^8 is also 81.
Now, to find the constant term in the expansion of (3x^2 + 1/x)^8, we need to multiply the constant terms obtained from both terms:
Constant term = (81) * (81) = 6561
Therefore, the constant term in the expansion of (3x^2 + 1/x)^8 is 6561.
To find the constant term of the expansion of (3x²+(1/x))^8, we need to expand the expression using the binomial theorem.
The binomial theorem states that the expansion of (a + b)^n can be found using the formula:
(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
Where C(n,k) is the binomial coefficient, given by the formula:
C(n,k) = n! / (k! * (n-k)!)
In our case, the expression is (3x²+(1/x))^8.
First, let's expand the expression using the binomial theorem:
(3x² + (1/x))^8 = C(8,0) * (3x²)^8 * (1/x)^0 + C(8,1) * (3x²)^7 * (1/x)^1 + C(8,2) * (3x²)^6 * (1/x)^2 + ... + C(8,7) * (3x²)^1 * (1/x)^7 + C(8,8) * (3x²)^0 * (1/x)^8
Now, simplify each term:
=> C(8,0) * (3^8 * x^(2*8)) * 1^0 + C(8,1) * (3^7 * x^(2*7)) * (1/x)^1 + C(8,2) * (3^6 * x^(2*6)) * (1/x)^2 + ... + C(8,7) * (3^1 * x^(2*1)) * (1/x)^7 + C(8,8) * (3^0 * x^(2*0)) * (1/x)^8
=> C(8,0) * 3^8 * x^16 + C(8,1) * 3^7 * x^14 * (1/x) + C(8,2) * 3^6 * x^12 * (1/x)^2 + ... + C(8,7) * 3^1 * x^2 * (1/x)^7 + C(8,8) * 3^0 * x^0 * (1/x)^8
Now, let's simplify further:
=> C(8,0) * 3^8 * x^16 + C(8,1) * 3^7 * x^14 * 1/x + C(8,2) * 3^6 * x^12 * 1/x^2 + ... + C(8,7) * 3^1 * x^2 * 1/x^7 + C(8,8) * 3^0 * x^0 * 1/x^8
Since we are interested in finding the constant term, we want x to have a power of 0. Looking at the terms, we find that the only term with x^0 is C(8,8) * 3^0 * x^0 * 1/x^8.
=> C(8,8) * 3^0 * x^0 * 1/x^8
Since the term does not have any x variables, we can simplify it further:
=> C(8,8) * 1 * 1/x^8
Now, let's calculate C(8,8):
C(8,8) = 8! / (8! * (8-8)!) = 1 / (1 * 1!) = 1
Substituting this back into our expression:
=> 1 * 1 * 1/x^8
=> 1 / x^8
Therefore, the constant term of the expansion of (3x² + (1/x))^8 is 1 / x^8.