To find the coefficient of the term containing x^16 in the expansion of (x^2 + 3)^10, you need to use the Binomial Theorem. The Binomial Theorem states that for any positive integer n, the expansion of (a + b)^n can be written as the sum of the terms of the form C(n, k) * a^(n-k) * b^k, where C(n, k) represents the binomial coefficient and is calculated using the formula:
C(n, k) = n! / (k! * (n - k)!)
In your case, the expression (x^2 + 3)^10 suggests that a = x^2 and b = 3, with n = 10. To find the coefficient of the term containing x^16, we need to determine the value of k in the expression C(10, k) * (x^2)^(10-k) * 3^k, where the exponent of x is 16. So, we can set up the equation:
10 - k = 16
Solving this equation gives us k = -6, but since k cannot be negative, there is no term with x^16 in the expansion (x^2 + 3)^10. Therefore, the coefficient is 0.
In summary, the coefficient of the term containing x^16 in the expansion of (x^2 + 3)^10 is 0.