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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\$$.

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\$$.
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\$$.