Proof Techniques Math Example 4

Follow the full solution, then compare it with the other examples linked below.

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Prove using mathematical induction:

3^n > 2n+1

for all

n \ge 2

1
Base case $n=2$ : $3^2 = 9 > 5 = 2(2)+1$ . True.
2
Inductive hypothesis: assume $3^k > 2k+1$ for some $k \ge 2$ .
3
Inductive step: $3^{k+1} = 3 \cdot 3^k > 3(2k+1) = 6k+3$ . Need to show $6k+3 > 2(k+1)+1 = 2k+3$ .
4
Indeed $6k+3 > 2k+3 \Leftrightarrow 4k > 0$ , which holds for $k \ge 2$ . Therefore $3^{k+1} > 2(k+1)+1$ . $\square$

Answer

3^n > 2n+1 \text{ for all } n \ge 2 \quad \square

Induction is natural here because the statement is indexed by

n

and involves a recursive structure (

3^{k+1} = 3 \cdot 3^k

). The key step is chaining the hypothesis through the factor of 3.

About Proof Techniques

Proof techniques are standard strategies for establishing mathematical claims under different structures.

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