True, but why?

$The\quad factorials\quad can\quad be\quad simplified\quad to:\quad 3\cdot 2+3\cdot 4+...+(k+1)\cdot k+(k+1)\cdot (k+2)+...+99\cdot 98+100\cdot 99\quad (k\quad is\quad even)\\ Grouping\quad and\quad factoring\quad terms:\quad \quad 3(2+4)+...+(k+1)(k+k+2)+...+99(100+98)\\ Operating:\quad 3(6)+...+(k+1)(2k+2)+...+99(198)=3(2\cdot 3)+...+(k+1)(2\cdot (k+1))+...+99(2\cdot 99)\\ Which\quad is\quad equal\quad to:\quad 2\cdot { 3 }^{ 2 }+...+2\cdot { (k+1) }^{ 2 }+...+2\cdot { 99 }^{ 2 }\\ Factoring:\quad 2({ 3 }^{ 2 }+...+{ (k+1) }^{ 2 }+...+{ 99 }^{ 2 })=\frac { 2(99\cdot 100\cdot 101) }{ 6 } -2=\frac { 2(98\cdot 100\cdot 102+100\cdot 101-98\cdot 100) }{ 6 } -2\\ =\frac { 16(49\cdot 50\cdot 51)+600 }{ 6 } -2=\frac { 16(49\cdot 50\cdot 51) }{ 6 } +98$