− 1 + 3 + 7 + 11 + 15 + ⋯ + ( 2 n + 3 ) = 1 2 ( n + 1 ) ( n + 4 ) {\displaystyle -1+3+7+11+15+\cdots +(2n+3)={\frac {1}{2}}(n+1)(n+4)}
L : ( 2 n + 3 ) = ( 2 ∗ 2 + 3 ) = 7 → − 1 + 3 + 7 = 9 R : 1 2 ( n + 1 ) ( n + 4 ) = 1 2 ( 2 + 1 ) ( 2 + 4 ) = 9 9 = 9 {\displaystyle {\begin{aligned}&L:(2n+3)=(2*2+3)=7\rightarrow -1+3+7=9\\&R:{\frac {1}{2}}(n+1)(n+4)={\frac {1}{2}}(2+1)(2+4)=9\\&9=9\\\end{aligned}}}
− 1 + 3 + 7 + 11 + 15 + ⋯ + ( 2 k + 3 ) = 1 2 ( k + 1 ) ( k + 4 ) {\displaystyle -1+3+7+11+15+\cdots +(2k+3)={\frac {1}{2}}(k+1)(k+4)}
− 1 + 3 + 7 + 11 + 15 + ⋯ + ( 2 k + 3 ) ⏟ = 1 2 ( k + 1 ) ( k + 4 ) + ( 2 k + 7 ) = 1 2 ( k + 1 ) ( k + 4 ) 1 2 ( k + 1 ) ( k + 4 ) + ( 2 k + 7 ) = 1 2 ( k + 1 ) ( k + 4 ) ( k + 1 ) ( k + 4 ) + 2 ( 2 k + 7 ) = ( k + 3 ) ( k + 6 ) k 2 + 5 k + 4 + 4 k + 14 = k 2 + 9 k + 18 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {-1+3+7+11+15+\cdots +(2k+3)} _{={\frac {1}{2}}(k+1)(k+4)}+(2k+7)={\frac {1}{2}}(k+1)(k+4)\\&{\frac {1}{2}}(k+1)(k+4)+(2k+7)={\frac {1}{2}}(k+1)(k+4)\\&(k+1)(k+4)+2(2k+7)=(k+3)(k+6)\\&k^{2}+5k+4+4k+14=k^{2}+9k+18\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
