משפט:: סכום מספרים אי-זוגיים ((Defines the sum of odd integers) : ∑ i = 1 n 2 i − 1 = n 2 {\displaystyle \sum _{i=1}^{n}2i-1=n^{2}}
בדיקה: ∑ i = 1 1 2 i − 1 = 2 − 1 = 1 ⇔ 1 2 {\displaystyle \sum _{i=1}^{1}2i-1=2-1=1\Leftrightarrow 1^{2}}
נניח כי הטענה נכונה לכל מספר טבעי : ∑ i = 1 n 2 i − 1 = n 2 {\displaystyle \sum _{i=1}^{n}2i-1=n^{2}}
נוכיח נכונותה: ∑ i = 1 n + 1 2 i − 1 = ( n + 1 ) 2 ∑ i = 1 n 2 i − 1 + [ 2 ( n + 1 ) − 1 ] = ( n + 1 ) 2 ∑ i = 1 n 2 i − 1 ⏟ n 2 + 2 ( n + 1 ) − 1 = ( n + 1 ) 2 n 2 + 2 ( n + 1 ) − 1 = ( n + 1 ) 2 n 2 + 2 n + 2 − 1 = ( n + 1 ) 2 {\displaystyle {\begin{aligned}\sum _{i=1}^{n+1}2i-1=(n+1)^{2}\\\sum _{i=1}^{n}2i-1+[2(n+1)-1]=(n+1)^{2}\\\underbrace {\sum _{i=1}^{n}2i-1} _{n^{2}}+2(n+1)-1=(n+1)^{2}\\n^{2}+2(n+1)-1=(n+1)^{2}\\n^{2}+2n+2-1=(n+1)^{2}\end{aligned}}}
∑ i = 1 n 2 i − 1 = n 2 {\displaystyle {\begin{aligned}\sum _{i=1}^{n}2i-1=n^{2}\end{aligned}}}
![{\displaystyle {\begin{aligned}\sum _{i=1}^{n+1}2i-1=(n+1)^{2}\\\sum _{i=1}^{n}2i-1+[2(n+1)-1]=(n+1)^{2}\\\underbrace {\sum _{i=1}^{n}2i-1} _{n^{2}}+2(n+1)-1=(n+1)^{2}\\n^{2}+2(n+1)-1=(n+1)^{2}\\n^{2}+2n+2-1=(n+1)^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2578b042e06d68ca3f00fa22a10342bfa4716df)
