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Misalkan \$$f(x) = \\displaystyle\\sum_{n=2}^{\\infty} x^{n} \$$ deret pangkat dengan himpunan kekonvergenan \$$|x|<1 \$$.

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\$$f\\left( \\dfrac{1}{\\var{a}} \\right) = \$$[[0]].

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Misalkan \$$g\\left( \\dfrac{1}{x} \\right)=xf''(x) \$$.

\n

Maka nilai dari \$$g(\\var{a})= \$$ [[0]].

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Himpunan kekonvergenan deret \$$g(x) \$$ adalah $\\ldots$.

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Dengan menggunakan deret pangkat \$$g(x) \$$, tentukan $\\displaystyle\\sum_{n=2}^\\infty \\dfrac{n(n-1)}{\\simplify{{a}+1}^n} =$ [[0]].

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