An algebra problem by Wildan Bagus Wicaksono

$f(1) = \frac { 2017 }{ 1 }$

$f(1)+f(2)={ 2 }^{ 2 }f(2)\\ f(1)=3f(2)\\ \frac { 2017 }{ 3 } =f(2)$

$f(1)+f(2)+f(3)={ 3 }^{ 2 }f(3)\\ f(1)+f(2)=8f(3)\\ \frac { 2017\cdot 3+2017 }{ 3 } =8f(3)\\ \frac { 2017\cdot 4 }{ 3\cdot 8 } =f(3)\\ \frac { 2017 }{ 6 } =f(3)$

$f(1)+f(2)+f(3)+f(4)={ 4 }^{ 2 }f(4)\\ f(1)+f(2)+f(3)=15f(4)\\ \frac { 2017\cdot 6+2017\cdot 2+2017 }{ 6 } =15f(4)\\ \frac { 2017\cdot 9 }{ 6\cdot 15 } =f(4)\\ \frac { 2017 }{ 10 } =f(4)$

$...$

$f(1)+f(2)+f(3)+...+f(n)={ n }^{ 2 }f(n)$ Notice the denominator of each fraction. The numbers form rows of 1, 3, 6, 10, .... The form of the sequence is the arithmetic sequence of level 2.

${ U }_{ n }=a+(n-1)b+\frac { (n-1)(n-2)c }{ 2! }$

Thus the denominator of $f (2017)$ ,

${ U }_{ 2017 }=1+(2017-1)2+\frac { (2017-1)(2017-2)1 }{ 2 } \\ { U }_{ 2017 }=1+2016\cdot 2+\frac { 2016\cdot 2015 }{ 2 } \\ { U }_{ 2017 }=1+2016\cdot 2+1008\cdot 2015\\ { U }_{ 2017 }=1+1008(2\cdot 2+2015)\\ { U }_{ 2017 }=1+1008\cdot 2019\\ { U }_{ 2017 }=1+1008(2018+1)\\ { U }_{ 2017 }=1+1008\cdot 2018+1008\\ { U }_{ 2017 }=1009+1008\cdot 2018\\ { U }_{ 2017 }=1009(1+1008\cdot 2)\\ { U }_{ 2017 }=1009(1+2016)\\ { U }_{ 2017 }=1009\cdot 2017$

So,

$f(2017)=\frac { 2017 }{ 1009\cdot 2017 } \\ f(2017)=\frac { 1 }{ 1009 } \\ 2018\cdot f(2017)=2018\cdot \frac { 1 }{ 1009 } \\ 2018\cdot f(2017)=2$