Variable Summations

$If\quad we\quad make\quad { S }_{ 1 }=f(x),\quad then\quad \frac { { S }_{ 1 }^{ \quad 2 } }{ x } ={ S }_{ 1 }\\ \\ Hence:\quad { { S }_{ 1 } }^{ 2 }-x{ S }_{ 1 }=0\quad ;\quad { S }_{ 1 }({ S }_{ 1 }-x)=0\quad \\ [and\quad we\quad assume\quad { S }_{ 1 }\neq 0\quad \\ (as\quad unless\quad x=0\quad it\quad must\quad have\quad value)]\\ { S }_{ 1 }=x.\\ Now\quad we\quad need\quad to\quad find\quad a\quad general\quad formula\quad for\quad \\ \sum _{ x }^{ n }{ f(x) } \quad where\quad 'n'\quad is\quad just\quad an\quad integer\quad greater\quad than\quad x.\\ As\quad f\left( x \right) =x\quad ,\quad \sum _{ x }^{ n }{ f\left( x \right) } =\quad x+(x+1)+(x+2)...+n\quad .\quad \\ we\quad can\quad say\quad \sum _{ x }^{ n }{ f\left( x \right) } =\quad (n-1)x\quad +\quad \frac { 1 }{ 2 } (n-x)(n-x+1)\quad ;\quad \\ \sum _{ x }^{ n }{ f\left( x \right) } =\quad nx-x\quad +\quad \frac { 1 }{ 2 } { n }^{ 2 }+n-2nx+{ x }^{ 2 }-x\quad ;\quad \\ This\quad simplifies\quad to:\quad \sum _{ x }^{ n }{ f\left( x \right) } =\frac { { x }^{ 2 }-3x+{ n }^{ 2 }+n }{ 2 } \\ \\ Now\quad we\quad must\quad try\quad to\quad find\quad 'x':\\ \\ If\quad g\left( x \right) =\sqrt { x+\sqrt { x+\sqrt { x+\sqrt { x+\sqrt { x+\sqrt { x+\sqrt { x } ... } } } } } } \\ we\quad can\quad say\quad that\quad { { S }_{ 2 } }^{ 2 }-x={ S }_{ 2 };\\ \\ This\quad can\quad be\quad solved\quad (as\quad any\quad quadratic)\quad to\quad give:\quad \\ S=\frac { 1\pm \sqrt { 1+4x } }{ 2 } .\\ We\quad are\quad told\quad that\quad g(x)=\left| x \right| \quad hence\quad \frac { 1\pm \sqrt { 1+4x } }{ 2 } =\left| x \right| \\ This\quad can\quad be\quad solved\quad in\quad two\quad unique\quad ways\quad \\ (depending\quad on\quad signs),\quad simply\quad by\quad rearrangement:\\ One\quad way\quad yields\quad a\quad single\quad result\quad x=0,\\ the\quad other\quad yields\quad two:\quad x=0\quad and\quad x=2.\\ We\quad know\quad x\neq 0\quad from\quad the\quad question,\\ hence\quad for\quad this\quad solution\quad x=2.\\ \\ Therefore\quad \sum _{ 5x }^{ { (5x) }^{ 2 } }{ f\left( x \right) } can\quad be\quad changed\quad to\quad \\ \sum _{ 10 }^{ 100 } f\left( x \right) \quad by\quad input\quad of\quad x=2.\\ We\quad know\quad that\quad the\quad summation\quad equation\quad is:\\ \quad \frac { { x }^{ 2 }-3x+{ n }^{ 2 }+n }{ 2 } \\ \quad \{ or\quad \frac { { 2x }^{ 2 }-3x+{ x }^{ 4 } }{ 2 } \} .\\ Hence\quad to\quad find\quad the\quad value\quad of\quad the\quad summation:\quad \\ \\ \frac { { 2(100) }^{ 2 }-3(100)+{ (100) }^{ 4 } }{ 2 } -\frac { { 2(9) }^{ 2 }-3(9)+{ (9) }^{ 4 } }{ 2 } \\ =\quad 50009850-3348\\ \quad =\quad \boxed { 50006502 } .$

P.S. Apologies for the horrendous layout of this solution. :/