1 1 + 2 2 + 3 2 + 4 2 + ⋯ + ( 2 n ) 2 = n 3 ( 2 n + 1 ) ( 4 n + 1 ) {\displaystyle 1^{1}+2^{2}+3^{2}+4^{2}+\cdots +(2n)^{2}={\frac {n}{3}}(2n+1)(4n+1)}
L : ( 2 n ) 2 = 2 2 → 1 2 + 2 2 = 5 R : n 3 ( 2 n + 1 ) ( 4 n + 1 ) = 1 3 ( 2 + 1 ) ( 4 + 1 ) = 5 5 = 5 {\displaystyle {\begin{aligned}&L:(2n)^{2}=2^{2}\rightarrow 1^{2}+2^{2}=5\\&R:{\frac {n}{3}}(2n+1)(4n+1)={\frac {1}{3}}(2+1)(4+1)=5\\&5=5\\\end{aligned}}}
1 1 + 2 2 + 3 2 + 4 2 + ⋯ + ( 2 k ) 2 = k 3 ( 2 k + 1 ) ( 4 k + 1 ) {\displaystyle 1^{1}+2^{2}+3^{2}+4^{2}+\cdots +(2k)^{2}={\frac {k}{3}}(2k+1)(4k+1)}
1 1 + 2 2 + 3 2 + 4 2 + ⋯ + ( 2 k ) 2 ⏟ = k 3 ( 2 k + 1 ) ( 4 k + 1 ) + ( 2 k + 1 ) 2 + ( 2 k + 2 ) 2 = k + 1 3 ( 2 k + 3 ) ( 4 k + 5 ) k 3 ( 2 k + 1 ) ( 4 k + 1 ) + ( 2 k + 1 ) 2 + ( 2 k + 2 ) 2 = k + 1 3 ( 2 k + 3 ) ( 4 k + 5 ) k ( 2 k + 1 ) ( 4 k + 1 ) + 3 ( 2 k + 1 ) 2 + 3 ( 2 k + 2 ) 2 = ( k + 1 ) ( 2 k + 3 ) ( 4 k + 5 ) k ( 8 k 2 + 6 k + 1 ) + 3 ( 4 k 2 + 4 k + 1 ) + 3 ( 4 k 2 + 4 k + 1 ) = ( 2 k 2 + 5 k + 3 ) ( 4 k + 5 ) 8 k 3 + 6 k 2 + k + 12 k 2 + 12 k + 3 + 12 k 2 + 24 k + 12 = 8 k 3 + 10 k 2 + 20 k 2 + 25 k + 12 k + 15 30 k 2 + 15 + 37 k = 30 k 2 + 15 + 37 k 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {1^{1}+2^{2}+3^{2}+4^{2}+\cdots +(2k)^{2}} _{={\frac {k}{3}}(2k+1)(4k+1)}+(2k+1)^{2}+(2k+2)^{2}={\frac {k+1}{3}}(2k+3)(4k+5)\\&{\frac {k}{3}}(2k+1)(4k+1)+(2k+1)^{2}+(2k+2)^{2}={\frac {k+1}{3}}(2k+3)(4k+5)\\&k(2k+1)(4k+1)+3(2k+1)^{2}+3(2k+2)^{2}=(k+1)(2k+3)(4k+5)\\&k(8k^{2}+6k+1)+3(4k^{2}+4k+1)+3(4k^{2}+4k+1)=(2k^{2}+5k+3)(4k+5)\\&8k^{3}+6k^{2}+k+12k^{2}+12k+3+12k^{2}+24k+12=8k^{3}+10k^{2}+20k^{2}+25k+12k+15\\&30k^{2}+15+37k=30k^{2}+15+37k\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
