Prove $\sum^n_{i=1} (2i-1)=n^2$ by induction

Question

The problem is to prove that $$\sum^n_{i=1} (2i-1)=n^2$$ for all $n \geq 1$ by induction.

Answer 1

If $n=1$, then $\sum_{i=1}^{n}(2i-1) = 2-1 = 1 = n^{2}$; if $n \geq 1$ and $\sum_{i=1}^{n}(2i-1) = n^{2}$, then $$ \sum_{i=1}^{n+1}(2i-1) = n^{2} + 2(n+1) - 1 = n^{2} + 2n + 1 = (n+1)^{2}; $$ by the principle of induction we are done.

Answer 2

First, show that this is true for $n=1$:

$\sum\limits_{i=1}^{1}2i-1=1^2$

Second, assume that this is true for $n$:

$\sum\limits_{i=1}^{n}2i-1=n^2$

Third, prove that this is true for $n+1$:

$\sum\limits_{i=1}^{n+1}2i-1=$

$\left(\color\red{\sum\limits_{i=1}^{n}2i-1}\right)+2(n+1)-1=$

$\color\red{n^2}+2(n+1)-1=$

$n^2+2n+1=$

$(n+1)^2$

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Please note that the assumption is used only in the part marked red.