expand (2+x)^5(1-2x)^6 as far as the term in x3

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StopLearn Team Staff answered 1 year ago

To expand the expression (2+x)^5(1-2x)^6 as far as the term in x^3, we need to perform a binomial expansion using the binomial theorem. The binomial theorem states that for any two binomials (a + b)^n and (c + d)^m, the expanded form can be expressed as a sum of terms, where each term is obtained by multiplying one term from the first binomial with one term from the second binomial.
(2+x)^5 = C(5,0)(2)^5(x)^0 + C(5,1)(2)^4(x)^1 + C(5,2)(2)^3(x)^2 + C(5,3)(2)^2(x)^3 + …
C(n,k) represents the binomial coefficient, also known as “n choose k”, which calculates the number of ways to choose k items from a set of n items. In this case, C(5,0) = 1, C(5,1) = 5, C(5,2) = 10, C(5,3) = 10, and the terms continue in the same pattern.
Next, we expand (1-2x)^6:
(1-2x)^6 = C(6,0)(1)^6(-2x)^0 + C(6,1)(1)^5(-2x)^1 + C(6,2)(1)^4(-2x)^2 + C(6,3)(1)^3(-2x)^3 + …
Now, let’s combine the two expansions:
(2+x)^5(1-2x)^6 = [(2+x)^5]*[(1-2x)^6]
= [C(5,0)(2)^5(x)^0 + C(5,1)(2)^4(x)^1 + C(5,2)(2)^3(x)^2 + C(5,3)(2)^2(x)^3 + …]

• [C(6,0)(1)^6(-2x)^0 + C(6,1)(1)^5(-2x)^1 + C(6,2)(1)^4(-2x)^2 + C(6,3)(1)^3(-2x)^3 + …]

To find the term involving x^3, we need to multiply terms from both expansions where the exponents of x add up to 3.
The term involving x^3 is obtained by multiplying the term C(5,1)(2)^4(x)^1 from the first expansion with the term C(6,2)(1)^4(-2x)^2 from the second expansion:
Term in x^3 = C(5,1)C(6,2)(2)^4*(x)^1*(1)^4*(-2x)^2
Simplifying this expression gives the term involving x^3 in the expansion of (2+x)^5(1-2x)^6.