INDUKSI MATEMATIKA 1^2 + 3^2 + 5^2 + ... + (2n + 1)^2 = (n + 1) (2n + 1) (2n + 3) / 3

Diketahui bahwa
[tex]{ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+...+{ (2n+1) }^{ 2 }=\frac { (n+1)(2n+1)(2n+3) }{ 3 } [/tex]
Pilih n = 1, sehingga 2n + 1 = 2 + 1 = 3.
[tex]{ 1 }^{ 2 }+{ 3 }^{ 2 }=\frac { (1+1)(2+1)(2+3) }{ 3 } \\ 1+9=\frac { 2\times 3\times 5 }{ 3 } \\ 10=10[/tex]
Akan kita buktikan bahwa pernyataan tersebut benar.
Pilih untuk (n + 1) 
[tex]{ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+...+{ (2n+1) }^{ 2 }+{ (2(n+1)+1) }^{ 2 }[/tex] = [tex]\frac { \left( \left[ n+1 \right] +1 \right) \left( 2\left[ n+1 \right] +1 \right) \left( 2\left[ n+1 \right] +3 \right) }{ 3 } [/tex]
[tex]\left( { 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+...+{ (2n+1) }^{ 2 } \right) +{ (2n+2+1) }^{ 2 }[/tex] = [tex]\frac { \left( n+2 \right) (2n+2+1)(2n+2+3) }{ 3 } [/tex]
[tex]\frac { (n+1)(2n+1)(2n+3) }{ 3 } +{ \left( 2n+3 \right) }^{ 2 }[/tex] =[tex]\frac { (n+2)(2n+3)(2n+5) }{ 3 } [/tex]
[tex]{ 4n }^{ 2 }+2\cdot 2n\cdot 3+9 =[tex]\frac { (n+2)(2n+3)(2n+5) }{ 3 } -\frac { (n+1)(2n+1)(2n+3) }{ 3 } [/tex] [/tex]
[tex]{ 4n }^{ 2 }+12n+9=\frac { (n+2)(2n+3)(2n+5)-(n+1)(2n+1)(2n+3) }{ 3 } [/tex]
[tex]3({ 4n }^{ 2 }+12n+9)=\left( 2n+3 \right) \left( \left[ n+2 \right] \left[ 2n+5 \right] -\left[ n+1 \right] \left[ 2n+1 \right] \right) [/tex][tex]
{ 12n }^{ 2 }+36n+27=(2n+3)\left( \left[ { 2n }^{ 2 }+9n+10 \right] -\left[ { 2n }^{ 2 }+3n+1 \right] \right) [/tex]
[tex]{ 12n }^{ 2 }+36n+27=(2n+3)({ 2n }^{ 2 }+9n+10-{ 2n }^{ 2 }-3n-1)\\ { 12n }^{ 2 }+36n+27=(2n+3)(6n+9)[/tex]
[tex]{ 12n }^{ 2 }+36n+27={ 12n }^{ 2 }+36n+27[/tex]
Jadi, terbukti bahwa [tex]{ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+...+{ (2n+1) }^{ 2 }=\frac { (n+1)(2n+1)(2n+3) }{ 3 } [/tex]