This is surely harder than ever !!!!!!

Given expression = $f(19)+\sum_{x=20}^{101}f(x)$

$=f(19)+\sum_{x=10}^{50}[f(2x)+f(2x+1)]$

$=94+\sum_{x=10}^{50}(2x+1)^2$

$=94+\sum_{x=1}^{50}(2x+1)^2-\sum_{x=1}^{9}(2x+1)^2$

$=175615$

I used the method of Sujoy Roy, but I came up with the following expression from scratch:

$\sum _{x=0} ^{n} {(2x+1)^2} = \dfrac {(n+1)(2n+1)(2n+3)} {3}$

Obviously, the value of $\sum _{x=0} ^{n} {(2x+1)^2}$ can be obtained using computer.

The expression:

$S = f(19)+f(20)+f(21)+...+f(101)$

$\quad = f(19)+(f(20)+f(21)) +(f(22)+f(23)) +...+(f(100)+f(101))$

$\quad = 94 + 21^2 + 23^2 + ... + 101^2$

$\quad = 94 + \sum _{x=0} ^{50} {(2x+1)^2} - \sum _{x=0} ^{9} {(2x+1)^2}$

$\quad = 94 + \dfrac {(51)(101)(103)} {3} - \dfrac {(10)(19)(21)} {3}$

$\quad = 94 + 176851 - 1330 = \boxed{175615}$

What I did was really rushy.

Take the alternating sum

$(f(19) + f(20)) - (f(20)+f(21))+ ... - (f(100)+f(101)) = f(19) - f(101)$

$20^{2} - 21^{2} + ... + 100^{2} - 101^{2} = 94 - f(101)$

Solve for $f(101) = 5055$

Take all the sum

$f(19) + \sum_{x=20}^{101} (f(x)+f(x-1)) + f(101) = 94 + \sum_{x=20}^{101} x^{2} + 5055 = 351230$

Since all the $f(x)$ from $x=19$ to $101$ are repeated twice, so we get $\displaystyle \frac{351230}{2} = \boxed{175615}$

f is defined recursively.

I haven't taken the trouble to memoise the function for efficiency.

C-Code

$\text{\#include<stdio.h>}$

int f(int x)

{

if(x==19)

return(94);

return (x*x-f(x-1));

}

int main()

{

int x,sum=0;

for(x=19;x<=101;x++)

sum+=f(x);

printf("%d",sum);

return 0;

}

f(x) can be x(x+1)/2 +96(-1)^x Hope You can do the sum!