Countries I have visited:
∑ p = 1 n p 2 = 1 2 + 2 2 + 3 2 + . . . + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 {\displaystyle \sum _{p=1}^{n}p^{2}=1^{2}+2^{2}+3^{2}+...+n^{2}={\frac {n(n+1)(2n+1)}{6}}}
∑ p = 1 n ( p 2 ) + ( n + 1 ) 2 = ∑ p = 1 n + 1 p 2 {\displaystyle \sum _{p=1}^{n}(p^{2})+(n+1)^{2}=\sum _{p=1}^{n+1}p^{2}}
∑ p = 1 n + 1 p 2 = ( n + 1 ) ( n + 2 ) ( 2 ( n + 1 ) + 1 ) 6 = ( n 2 + 3 n + 2 ) ( 2 n + 3 ) 6 {\displaystyle \sum _{p=1}^{n+1}p^{2}={\frac {(n+1)(n+2)(2(n+1)+1)}{6}}={\frac {(n^{2}+3n+2)(2n+3)}{6}}}
∑ p = 1 n ( p 2 ) + ( n + 1 ) 2 = 2 n 3 + 9 n 2 + 13 n + 6 6 {\displaystyle \sum _{p=1}^{n}(p^{2})+(n+1)^{2}={\frac {2n^{3}+9n^{2}+13n+6}{6}}}
Q . E . D . {\displaystyle \ Q.E.D.}