- #1
zbobet2012
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I understand how Binomial expansion works, but I don't understand how to solve this problem.
Give the term of (2/x^2-x)^6 that has no x.
Give the term of (2/x^2-x)^6 that has no x.
The formula for binomial expansion is (a + b)ⁿ = ∑ (nCr) a^(n-r) b^r, where n is the exponent, a and b are the terms of the binomial, and nCr is the combination coefficient.
To solve a binomial expansion with no x term, you can use the formula (a + b)ⁿ = aⁿ + (nC1) a^(n-1) b + (nC2) a^(n-2) b² + ... + (nCn-1) a b^(n-1) + bⁿ. In this case, the term (2/x²) has an exponent of 6 and no x term, so it can be rewritten as (2)⁶ (x²)^-6 = 64/x¹².
The value of (nCr) in binomial expansion is the combination coefficient, which represents the number of ways to choose r objects from a set of n objects. It can be calculated using the formula (nCr) = n! / (r! (n-r)!).
There are 7 terms in a binomial expansion with an exponent of 6. This is because the formula (a + b)ⁿ has n+1 terms, and in this case, n is 6, so there are 7 terms.
Yes, you can use binomial expansion to solve equations with variables in the exponent. In this case, the equation (2/x^2-x)^6 has a variable in the exponent, and it can be simplified using the binomial expansion formula.