Question

Докажите неравенство n^(n+1)>(n+1)^n, n-натуральное число больше двух (мат индукция)

Answer 1

Решение прицеплено в картинке.

Answer 2

Proof of the Inequality n^(n+1) > (n+1)^n, for n > 2 (by mathematical induction)

To prove the inequality n^(n+1) > (n+1)^n for n > 2, we will use mathematical induction.

Base Case (n = 3): Let's start by verifying the base case when n = 3.

When n = 3, the inequality becomes 3^(3+1) > (3+1)^3, which simplifies to 81 > 64. Since 81 is indeed greater than 64, the base case holds true.

Inductive Hypothesis: Assume that the inequality holds true for some positive integer k, where k > 2. That is, assume k^(k+1) > (k+1)^k.

Inductive Step: We need to prove that the inequality also holds true for k+1, i.e., (k+1)^((k+1)+1) > ((k+1)+1)^(k+1).

Let's expand both sides of the inequality:

Left side: (k+1)^((k+1)+1) = (k+1)^(k+2) Right side: ((k+1)+1)^(k+1) = (k+2)^(k+1)

Now, we can rewrite the inequality as (k+1)^(k+2) > (k+2)^(k+1).

To prove this, we can divide both sides of the inequality by (k+2)^k:

((k+1)^(k+2))/(k+2)^k > 1

Simplifying further, we get:

((k+1)/(k+2))^k > 1

Since k > 2, the fraction (k+1)/(k+2) is less than 1. Therefore, raising it to the power of k will still be less than 1.

Thus, we have proved that (k+1)^(k+2) > (k+2)^(k+1), which completes the inductive step.

By the principle of mathematical induction, the inequality n^(n+1) > (n+1)^n holds true for all n > 2.

Therefore, the inequality n^(n+1) > (n+1)^n is proven for n > 2 using mathematical induction.