Find the numerical value of the coefficient x^11 in the expansion of (x^2+1/x)^10 in the powers of x.
general term = term(r+1)
=C(10,r) (x^2)^(10-r) (1/x)^r
= C(10,r) x^(20 - 2r) x^(-r)
= C(10,r) x^(20 -3r)
so 20 -3r = 11
-3r = -9
r = 3
C(10,3) = 120
the term containing x^11 is 120x^11 and it is the 4th term
confirmation by Wolfram:
http://www.wolframalpha.com/input/?i=expand+%28x%5E2%2B1%2Fx%29%5E10
To find the numerical value of the coefficient of x^11 in the expansion of (x^2+1/x)^10, we can use the binomial theorem.
The binomial theorem states that for any real number a and b, and any positive integer n, the n-th power of the sum a + b can be expanded using the binomial coefficients. The coefficients are the numbers in front of each term in the expansion.
In our case, a = x^2 and b = 1/x. So we have (x^2 + 1/x)^10.
To find the coefficient of x^11, we need to find the term that contains x^11 in the expansion. This term is obtained by choosing the 11th power of x^2 in one term and the 0th power of 1/x in the other term.
Applying the binomial theorem, the coefficient of x^11 is given by the formula:
C(10, k) * (x^2)^(10-k) * (1/x)^k
where C(10, k) is the binomial coefficient, (x^2)^(10-k) represents the term with x^2 raised to the power of (10-k), and (1/x)^k represents the term with 1/x raised to the power of k.
In this case, we have k = 0 since we need to choose the 0th power of 1/x. Plugging these values into the formula, we have:
C(10, 0) * (x^2)^(10-0) * (1/x)^0
Simplifying further:
C(10, 0) * (x^2)^10 * 1
Since C(10, 0) = 1, we have:
1 * (x^2)^10 * 1
= (x^2)^10
= x^(2*10)
= x^20
Therefore, the numerical value of the coefficient of x^11 in the expansion of (x^2 + 1/x)^10 is 0, since there is no term with x^11 in the expansion.