Final answer:
The coefficient of x⁵ in the expansion (x+2)⁹ + x⁴(x+3)⁻ is found by using the binomial theorem to identify the appropriate terms in each binomial expansion that contribute to the x⁵ term and summing their coefficients. In this case, the coefficient from the first term is 715, and the coefficient from the second term is 3003, which sums up to 3718.
Step-by-step explanation:
The question asks for the coefficient of xⁱ in the expansion of (x+2)⁹ + x⁴(x+3)⁻.
We can use the binomial theorem which states that the expansion of (a+b)ⁿ includes terms like a²⁴b¹, a⁵b², and so forth, where the sum of the exponents in each term adds up to n. In this case, we are looking for the term that includes x⁵, which will come from the x term in the parentheses raised to the ninth power multiplied by the constant term raised to the remaining power such that the sum of the powers is 13.
From the first term (x+2)⁹, the x coefficient for x⁵ will be given by the term in the binomial expansion that has x⁵, which can be found using the formula C(n,k) = n! / (k!(n-k)!), where C(n,k) is the binomial coefficient for the term xⁿ(2)⁴. This coefficient is 13! / (9!4!), which simplifies to 715.
In the term x⁴(x+3)⁻, the x term is already raised to the fourth power, so we need an additional power of 5 to have x⁵. According to the binomial theorem, the coefficient we seek in this term is C(15,5), which is 15! / (5!10!) and simplifies to 3003. However, since this coefficient is multiplied by x⁴, the whole term for x⁵ will be 3003 * x⁴, making the coefficient of x⁵ in the whole second term 3003.
The final coefficient for x⁵ in the entire expression is the sum of the coefficients from both terms, so we add 715 + 3003 to get a final answer of 3718.
Adding the coefficients from both parts of the expression gives us the final coefficient for x⁵, using the general principle xPxq = x(p+q). Therefore, the coefficient of x⁵ in the expansion is 3718.