2 ∗ 3 − 21 + 4 ∗ 5 − 4 ∗ 3 + ⋯ + n ( n + 1 ) − n ( n − 1 ) = n 2 ( n + 2 ) {\displaystyle 2*3-21+4*5-4*3+\cdots +n(n+1)-n(n-1)={\frac {n}{2}}(n+2)}
L : n ( n + 1 ) − n ( n − 1 ) = 2 ( 2 + 1 ) − 2 ( 2 − 1 ) = 6 − 2 = 4 → 2 ∗ 3 − 2 ∗ 1 = 4 R : n 2 ( n + 2 ) = 2 2 ( 2 + 2 ) = 4 4 = 4 {\displaystyle {\begin{aligned}&L:n(n+1)-n(n-1)=2(2+1)-2(2-1)=6-2=4\rightarrow 2*3-2*1=4\\&R:{\frac {n}{2}}(n+2)={\frac {2}{2}}(2+2)=4\\&4=4\\\end{aligned}}}
2 ∗ 3 − 21 + 4 ∗ 5 − 4 ∗ 3 + ⋯ + k ( k + 1 ) − k ( k − 1 ) = k 2 ( k + 2 ) {\displaystyle 2*3-21+4*5-4*3+\cdots +k(k+1)-k(k-1)={\frac {k}{2}}(k+2)}
2 ∗ 3 − 21 + 4 ∗ 5 − 4 ∗ 3 + ⋯ + k ( k + 1 ) − k ( k − 1 ) ⏟ = k 2 ( k + 2 ) + ( k + 2 ) ( k + 3 ) − ( k + 2 ) ( k + 1 ) = k + 2 2 ( k + 4 ) k 2 ( k + 2 ) + ( k + 2 ) ( k + 3 ) − ( k + 2 ) ( k + 1 ) = k + 2 2 ( k + 4 ) k ( k + 2 ) + 2 ( k 2 + 5 k + 6 ) − 2 ( k 2 + 3 k + 2 ) = k 2 + 6 k + 8 k 2 + 2 k + 2 k 2 + 10 k + 12 − 2 k 2 − 6 k − 4 = k 2 + 6 k + 8 6 k + 8 = 6 k + 8 {\displaystyle {\begin{aligned}&\underbrace {2*3-21+4*5-4*3+\cdots +k(k+1)-k(k-1)} _{={\frac {k}{2}}(k+2)}+(k+2)(k+3)-(k+2)(k+1)={\frac {k+2}{2}}(k+4)\\&{\frac {k}{2}}(k+2)+(k+2)(k+3)-(k+2)(k+1)={\frac {k+2}{2}}(k+4)\\&k(k+2)+2(k^{2}+5k+6)-2(k^{2}+3k+2)=k^{2}+6k+8\\&k^{2}+2k+2k^{2}+10k+12-2k^{2}-6k-4=k^{2}+6k+8\\&6k+8=6k+8\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
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