Proof: By Induction
(related to Proposition: Inequality between Square Numbers and Powers of 2)
We provide a proof by induction for all natural numbers n\in\mathbb N.
- Base case: n=0
0^2 < 2^0=1.
- Induction step n\to n+1
- Assume, n^2 < 2 ^n is correct for an n\ge 0.
- Then \begin{align}(n+1)^2&=n^2+2n+1\nonumber\\
& \le n^2+n^2\nonumber\\
&=2\cdot n^2.\nonumber\\
& < 2\cdot 2^n\quad\text{(by assumption)}\nonumber\\
&=2^{n+1}\nonumber
\end{align}
- Thus, the sum of consecutive odd numbers from 1 to n equals the square of n.
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