Given the expression (2/x^2-x)^15
1. Find the number of teams in the expression.
2. Find the constant term
To determine the number of terms in the expression, we need to expand and simplify it.
To do this, we start by expanding the binomial using the binomial theorem.
The binomial theorem states that for any positive integer n,
(x + y)^n = (nC0)x^n(y^0) + (nC1)x^(n-1)(y^1) + ... + (nCn)y^n.
In this case, x is equal to 2/x^2 and y is equal to -x.
Applying the binomial theorem, we have:
(2/x^2 - x)^15 = (15C0)(2/x^2)^15(-x)^0 + (15C1)(2/x^2)^14(-x)^1 + ... + (15C15)(2/x^2)^0(-x)^15.
Simplifying, we get:
(2/x^2 - x)^15 = (15C0)(2^15/x^30) + (15C1)(2^14/x^28)(-x) + ... + (15C15)(1)(-x)^15.
Now, we can examine the form of each term to determine the number of terms and the constant term.
1. The binomial theorem shows that there are (15C0) + (15C1) + ... + (15C15) terms in the expansion. In this case, there are 16 terms.
2. The constant term occurs when all the x terms cancel out. Looking at the expression, the constant term occurs when the exponent of x in each term is 0. In this case, the constant term is given by:
(15C0)(2^15/x^30) = (2^15/x^30) = 2^15/x^30.
Therefore, the constant term is 2^15/x^30.