Final answer:
To prove the inequality an − bn ≤ nan−1 (a − b), we can use the binomial theorem to expand an and bn. By substituting these in the original inequality and using the identity (a^n - b^n) / (a - b) = a^(n-1) + a^(n-2)b + a^(n-3)b^2 + ... + ab^(n-2) + b^(n-1), we can see that the sum of terms on the left side is smaller than or equal to n(a^(n-1)). Thus, the inequality holds.
Step-by-step explanation:
To prove the inequality an − bn ≤ nan−1 (a − b), we can start by expanding an and bn using the binomial theorem. We have:
an = a^n
bn = b^n
Next, we substitute these into the original inequality:
a^n - b^n ≤ n(a^(n-1))(a - b)
a^n - b^n = (a^n - b^n) / (a - b) * (a - b)
Using the identity (a^n - b^n) / (a - b) = a^(n-1) + a^(n-2)b + a^(n-3)b^2 + ... + ab^(n-2) + b^(n-1), we can rewrite the inequality as:
a^(n-1) + a^(n-2)b + a^(n-3)b^2 + ... + ab^(n-2) + b^(n-1) ≤ n(a^(n-1))(a - b)
On the left side, we have a sum of terms, each of which is smaller than or equal to a^(n-1), since b < a. Therefore, the sum is also smaller than or equal to n(a^(n-1)).
Thus, we have proven the inequality an − bn ≤ nan−1 (a − b).