QuestionBeweisen Sie durch Induktion, dass $2n + 1 \leq n^2 \leq 2^n$ für alle $n \geq 4$ gilt.

Studdy Solution

STEP 1

Assumptions1. We are asked to prove the inequality $n+1 \leq n^{} \leq^{n}$ for all $n \geq4$ . . We will use the method of mathematical induction to prove this.

STEP 2

The principle of mathematical induction consists of two steps the base case and the inductive step.
First, we need to verify the base case. In this case, the base case is $n=4$ .
$2(4)+1 \leq4^{2} \leq2^{4}$

STEP 3

Calculate the values for the base case.
$9 \leq16 \leq16$

STEP 4

The base case holds true, so we can proceed to the inductive step.
Assume the inequality holds for some $n=k$ , i.e., $2k+1 \leq k^{2} \leq2^{k}$

STEP 5

We need to show that the inequality also holds for $n=k+1$ , i.e., $2(k+1)+1 \leq (k+1)^{2} \leq2^{k+1}$

STEP 6

implify the left-hand side of the inequality.
$2k+3 \leq (k+1)^{2} \leq2^{k+1}$

STEP 7

Expand the middle term.
$2k+3 \leq k^{2}+2k+1 \leq2^{k+1}$

STEP 8

Since we assumed that $2k+1 \leq k^{2}$ , we know that $2k+3 \leq k^{2}+2k+1$ .

STEP 9

Now we need to show that $k^{2}+2k+ \leq2^{k+}$ .

STEP 10

Since we assumed that $k^{2} \leq2^{k}$ , we know that $k^{2}+2k+ \leq2^{k}+2k+$ .

STEP 11

We need to show that $^{k}+k+ \leq^{k+}$ .

STEP 12

implify the right-hand side of the inequality.
$2^{k}+2k+ \leq2 \cdot2^{k}$

STEP 13

Since $2k+ \leq2^{k}$ (from our assumption), we know that $2^{k}+2k+ \leq2 \cdot2^{k}$ .

STEP 14

So, we have shown that if the inequality holds for $n=k$ , it also holds for $n=k+$ .
Therefore, by the principle of mathematical induction, the inequality $2n+ \leq n^{2} \leq2^{n}$ holds for all $n \geq4$ .

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