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Fungsi \\( f(x)=\\dfrac{1}{1+x-x^2} \\) mempunyai Deret Maclaurin \\( \\displaystyle\\sum_{n=0}^\\infty (-1)^nf_nx^n\\) (pada daerah kekonvergenannya) dengan \\( f_n \\) menyatakan suku ke-$n$ dari barisan Fibonacci, yakni $f_1=1$, $f_2=1$, dan $f_n=f_{n-1}+f_{n-2}$ untuk \\( n\\geq 3\\).

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Polinom Taylor orde $6$ dari $f(x)$ di sekitar \\(x=\\dfrac{1}{2}\\) adalah

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\\(f(x)\\approx\\) [[0]] $+$ [[1]] \\(\\left( x-\\dfrac{1}{2} \\right)^2\\) $+$ [[2]] \\(\\left( x-\\dfrac{1}{2} \\right)^4\\) $+$ [[3]] \\(\\left( x-\\dfrac{1}{2} \\right)^6\\).

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Koefisien suku \\( \\left( x-\\dfrac{1}{2} \\right)^{40} \\) dari uraian Deret Taylor \\( f(x) \\) di sekitar \\( x=\\dfrac{1}{2} \\) berbentuk \\( \\dfrac{a^{10}}{b^{10}} \\) dengan \\( a\\) dan \\( b\\) bilangan bulat. Maka \\( a+b =\\ldots\\).

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Nilai dari \\( f^{(40)}\\left(\\dfrac{1}{2}\\right) = n!\\cdot \\dfrac{a^{10}}{b^{10}}\\) dengan \\(a\\), \\(b\\), dan \\(n\\) bilangan bulat. Maka \\(a+b+n=\\ldots\\).

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