Uočimo da se izraz na desnoj strani može zapisati kao
f ( x ) = 2 ( x + 2 ) 2 3 + ( x + 2 ) ( x − 2 ) 3 + ( x − 2 ) 2 3 . f(x)=\frac{2}{\sqrt[3]{(x+2)^2}+\sqrt[3]{(x+2)(x-2)}+\sqrt[3]{(x-2)^2}}. f ( x ) = 3 ( x + 2 ) 2 + 3 ( x + 2 ) ( x − 2 ) + 3 ( x − 2 ) 2 2 . Također, uočimo da je nazivnik oblika a 2 + a b + b 2 a^2+ab+b^2 a 2 + ab + b 2 a = x + 2 3 a=\sqrt[3]{x+2} a = 3 x + 2 b = x − 2 3 b=\sqrt[3]{x-2} b = 3 x − 2 a − b = x + 2 3 − x − 2 3 a-b=\sqrt[3]{x+2}-\sqrt[3]{x-2} a − b = 3 x + 2 − 3 x − 2
f ( x ) = 2 ( x + 2 ) 2 3 + ( x + 2 ) ( x − 2 ) 3 + ( x − 2 ) 2 3 ⋅ x + 2 3 − x − 2 3 x + 2 3 − x − 2 3 = 2 ( x + 2 3 − x − 2 3 ) ( x + 2 ) 3 3 − ( x − 2 ) 3 3 = 2 ( x + 2 3 − x − 2 3 ) ( x + 2 ) − ( x − 2 ) = 2 ( x + 2 3 − x − 2 3 ) 4 = x + 2 3 − x − 2 3 2 . \begin{aligned} f(x) &=\frac{2}{\sqrt[3]{(x+2)^2}+\sqrt[3]{(x+2)(x-2)}+\sqrt[3]{(x-2)^2}} \cdot\frac{\sqrt[3]{x+2}-\sqrt[3]{x-2}}{\sqrt[3]{x+2}-\sqrt[3]{x-2}}\\ &=\frac{2(\sqrt[3]{x+2}-\sqrt[3]{x-2})}{\sqrt[3]{(x+2)^3}-\sqrt[3]{(x-2)^3}}\\ &=\frac{2(\sqrt[3]{x+2}-\sqrt[3]{x-2})}{(x+2)-(x-2)} =\frac{2(\sqrt[3]{x+2}-\sqrt[3]{x-2})}{4}\\ &=\frac{\sqrt[3]{x+2}-\sqrt[3]{x-2}}{2}. \end{aligned} f ( x ) = 3 ( x + 2 ) 2 + 3 ( x + 2 ) ( x − 2 ) + 3 ( x − 2 ) 2 2 ⋅ 3 x + 2 − 3 x − 2 3 x + 2 − 3 x − 2 = 3 ( x + 2 ) 3 − 3 ( x − 2 ) 3 2 ( 3 x + 2 − 3 x − 2 ) = ( x + 2 ) − ( x − 2 ) 2 ( 3 x + 2 − 3 x − 2 ) = 4 2 ( 3 x + 2 − 3 x − 2 ) = 2 3 x + 2 − 3 x − 2 . Slijedi
f ( 4 ) + f ( 8 ) + f ( 12 ) + ⋯ + f ( 2024 ) = 1 2 ( 6 3 − 2 3 + 10 3 − 6 3 + 14 3 − 10 3 + ⋯ + 2026 3 − 2022 3 ) = 1 2 ( 2026 3 − 2 3 ) . \begin{aligned} f(4)+f(8)+f(12)+\cdots+f(2024) &=\frac12(\sqrt[3]{6}-\sqrt[3]{2}+\sqrt[3]{10}-\sqrt[3]{6}+\sqrt[3]{14}-\sqrt[3]{10}+\cdots+\sqrt[3]{2026}-\sqrt[3]{2022})\\ &=\frac12(\sqrt[3]{2026}-\sqrt[3]{2}). \end{aligned} f ( 4 ) + f ( 8 ) + f ( 12 ) + ⋯ + f ( 2024 ) = 2 1 ( 3 6 − 3 2 + 3 10 − 3 6 + 3 14 − 3 10 + ⋯ + 3 2026 − 3 2022 ) = 2 1 ( 3 2026 − 3 2 ) . 