Answer:
Explanation:
A) The Binomial Theorem states that for any positive integer n, the expansion of (1 + x)^n can be expressed as:
(1 + x)^n = C(n, 0) * 1^n * x^0 + C(n, 1) * 1^(n-1) * x^1 + C(n, 2) * 1^(n-2) * x^2 + ... + C(n, n-1) * 1^1 * x^(n-1) + C(n, n) * 1^0 * x^n
where C(n, k) represents the binomial coefficient, defined as C(n, k) = n! / (k! * (n - k)!), and "!" denotes the factorial operation.
Using the above formula, we can find the first four terms of the polynomial (1 + x)^n:
Term 1: C(n, 0) * 1^n * x^0 = 1 * 1 * 1 = 1
Term 2: C(n, 1) * 1^(n-1) * x^1 = n * x
Term 3: C(n, 2) * 1^(n-2) * x^2 = n * (n - 1) / 2 * x^2
Term 4: C(n, 3) * 1^(n-3) * x^3 = n * (n - 1) * (n - 2) / 6 * x^3
So the first four terms of the polynomial (1 + x)^n are: 1, n * x, n * (n - 1) / 2 * x^2, and n * (n - 1) * (n - 2) / 6 * x^3.
B) When x is less than 1, x^n becomes a smaller and smaller fraction as n increases. This is because each subsequent term in the expansion (1 + x)^n is multiplied by x, making it smaller compared to the previous term. Therefore, as n gets larger, x^n approaches 0 rapidly. In this case, the approximation of (1 + x)^n by 1 + nx becomes more accurate because the higher-order terms involving x^2, x^3, and so on, become negligible. Thus, 1 + nx becomes a good approximation of (1 + x)^n for small values of x. This is known as a linear approximation or a first-order approximation.