Find the 6th term in the expansion (x^2 - 1/x)^25
It is a Binomial Theorem problem and I am extremely confused. Please and thank you!
To find the 6th term in the expansion of (x^2 - 1/x)^25, you can use the Binomial Theorem. The Binomial Theorem states that the expansion of (a + b)^n can be written as follows:
(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, k) * a^(n-k) * b^k + ... + C(n, n) * a^0 * b^n
In this case, a = x^2 and b = -1/x. Let's calculate the 6th term step-by-step:
Step 1: Determine the values of n and k.
In this case, n = 25 and k = 6.
Step 2: Calculate the binomial coefficient.
C(n, k) = C(25, 6) = 25! / (6! * (25-6)!)
Step 3: Calculate the powers of a and b.
a^(n-k) = (x^2)^(25-6) = x^2^19 = x^38
b^k = (-1/x)^6 = (-1)^6 / x^6 = 1 / x^6
Step 4: Calculate the terms.
6th term = C(25, 6) * (x^2)^(25-6) * (-1/x)^6
= C(25, 6) * x^38 * 1 / x^6
= C(25, 6) * x^32
Therefore, the 6th term in the expansion of (x^2 - 1/x)^25 is C(25, 6) * x^32.
To find the 6th term in the expansion of (x^2 - 1/x)^25, we can use the Binomial Theorem. The Binomial Theorem provides a formula for expanding expressions of the form (a + b)^n, where a and b are any real numbers, and n is a positive integer.
The Binomial Theorem states that the coefficients of the terms in the expansion are given by the binomial coefficients, commonly denoted as nCr or C(n,r), where n is the exponent and r is the term number. The binomial coefficient nCr is calculated as nCr = n! / (r! * (n-r)!), where ! denotes factorial.
In our case, we have (x^2 - 1/x)^25. We want to find the 6th term in the expansion, which corresponds to r = 6.
To find the coefficient of the 6th term, we need to calculate the binomial coefficient C(25, 6). First, let's compute the factorial for each term:
25! = 25 * 24 * 23 * 22 * 21 * 20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
6! = 6 * 5 * 4 * 3 * 2 * 1
(25-6)! = 19! = 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
Now we can calculate the binomial coefficient:
C(25, 6) = 25! / (6! * (25-6)!) = (25 * 24 * 23 * 22 * 21 * 20) / (6 * 5 * 4 * 3 * 2 * 1 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)
Simplifying this expression will give us the coefficient of our 6th term.
Finally, to determine the term, we need to consider the exponents of x^2 and 1/x in the expansion. Since we want the 6th term, the exponent of x^2 will be (25 - 6) = 19, and the exponent of 1/x will be 6.
Therefore, the 6th term in the expansion of (x^2 - 1/x)^25 will be given by the coefficient we calculated multiplied by (x^2)^19 * (1/x)^6.
I hope this explanation helps you understand the process of finding the 6th term in the expansion of (x^2 - 1/x)^25 using the Binomial Theorem.
So this expansion has 26 terms.Let consider the (r+1)th term and let's name it T(r+1)
So T(r+1)= nCr*[(x^2)(n-r)]*[(1/x)^r
nCr=[n!]/[(n-r)!*(r)!]
= nCr*[(x^2)(n-r)]*[(x^-1)^r] = nCr*[(x)^(2n-2r-r)]=nCr*[(x)^(2n-3r)]
When r=5 ,we get T(6) which is the sixth term of the expansion.