An algebra problem by Fidel Simanjuntak

$\begin{aligned} f(x) + f(x+1) & = 2x^2 \\ \implies f(x+1) & = 2x^2 - f(x) \\ f({\color{#3D99F6}1}) & = - f(0) & = {\color{#3D99F6}1}({\color{#3D99F6}1}-1)+(-1)^{\color{#3D99F6}1}f(0) \\ f({\color{#3D99F6}2}) & = 2 - f(1) = 2+f(0) & = {\color{#3D99F6}2}({\color{#3D99F6}2}-1)+(-1)^{\color{#3D99F6}2}f(0) \\ f({\color{#3D99F6}3}) & = 6 - f(0) & = {\color{#3D99F6}3}({\color{#3D99F6}3}-1)+(-1)^{\color{#3D99F6}3}f(0) \\ f({\color{#3D99F6}4}) & = 12 + f(0) & = {\color{#3D99F6}4}({\color{#3D99F6}4}-1)+(-1)^{\color{#3D99F6}4}f(0) \\ \cdots & = \cdots \\ \implies f({\color{#3D99F6}n}) & = {\color{#3D99F6}n}({\color{#3D99F6}n}-1)+(-1)^{\color{#3D99F6}n}f(0) & \small \color{#3D99F6} \text{See proof below.} \end{aligned}$

Then we have:

$\begin{aligned} f(31) & = 99 \\ 31(30) - f(0) & = 99 \\ \implies f(0) & = 831 \\ \implies f(130) & = 130(129) + f(0) \\ & = \boxed{17601} \end{aligned}$

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Proof:

Let us proof by induction the claim that $f(n) = n(n-1)+(-1)^nf(0)$ is true for all $n \ge 1$ .

• For $n=1$ , $f(1) = 2(0) - f(0) = 1(1-1) + (-1)^1f(0)$ , therefore, the claim is true for $n=1$ .
• Assuming the claim is true for $n$ , then

$\begin{aligned} \quad f({\color{#3D99F6}n+1}) & = 2n^2 - f(n) \\ & = 2n^2 - \left(n(n-1)+(-1)^nf(0)\right) \\ & = n^2 + n +(-1)^{n+1}f(0) \\ & = (n+1)n +(-1)^{n+1}f(0) \\ & = ({\color{#3D99F6}n+1})({\color{#3D99F6}n+1}-1) +(-1)^{\color{#3D99F6}n+1}f(0) \end{aligned}$

$\quad$ Therefore, the claim is also true for $n+1$ and hence true for all $n\ge 1$ .