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第47页

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$解：设S=-1-2-3-4-···-199-200①，$

$则S=-200-199-198-197-···-2-1②.$

$由①+②，得2S=-201×200，$

$ 即2S=-40200，$

$所以S=-20100，$

$即-1-2-3-4-···-199-200=-20100$

解：令$S=1+5+5^2+5^3+...+5^{217}+5^{218}①$，

则$5S=5+5^2+5^3+5^4+···+5^{218}+5^{219}②.$

由②-①，得$5S-S=5^{219}-1$，

所以$S=\frac {5^{219}-1}{4}$，

即$1+5+5^2+5^3+···+5^{217}+5^{218}=\frac {5^{219}-1}{4}$

$解：设S=1+\frac {1}{2}+\frac {1}{2^2}+\frac {1}{2^3}+···+\frac {1}{2^{2025}}①，$

$则\frac {1}{2}S=\frac {1}{2}+\frac {1}{2^2}+\frac {1}{2^3}+\frac {1}{2^4}+···+\frac {1}{2^{2026}}②.$

$由①-②，得\frac {1}{2}S=1-\frac {1}{2^{2026}}，$

$所以S=2-\frac {1}{2^{2025}}，即1+\frac {1}{2}+\frac {1}{2^2}+\frac {1}{2^3}+...+\frac {1}{2^{2025}}=2-\frac {1}{2^{2025}}$

$解：(1)原式=\frac {1}{4}× (\frac {1}{3}-\frac {1}{7})+\frac {1}{4}× (\frac {1}{7}-\frac {1}{11})+\frac {1}{4} ×(\frac {1}{11}-\frac {1}{15})+···+\frac {1}{4}×(\frac {1}{55}-\frac {1}{59})$

$=\frac {1}{4}×(\frac {1}{3}-\frac {1}{7}+\frac {1}{7}-\frac {1}{11}+\frac {1}{11}-\frac {1}{15}+···+ \frac {1}{55}-\frac {1}{59})$

$=\frac {1}{4}×(\frac {1}{3}-\frac {1}{59})$

$=\frac {14}{177}$

$(2)原式=-\frac {1}{1×3}-\frac {1}{3×5}-\frac {1}{5×7}-\frac {1}{7×9}-\frac {1}{9×11}-\frac {1}{11×13}$

$=-\frac {1}{2}×(1-\frac {1}{3})-\frac {1}{2}×(\frac {1}{3}-\frac {1}{5})-\frac {1}{2}× (\frac {1}{5}-\frac {1}{7})-\frac {1}{2}×(\frac {1}{7}-\frac {1}{9})-\frac {1}{2}×(\frac {1}{9}-\frac {1}{11})-\frac {1}{2}×(\frac {1}{11}-\frac {1}{13})$

$=-\frac {1}{2}×(1-\frac {1}{3}+\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+\frac {1}{7}-\frac {1}{9}+\frac {1}{9}-\frac {1}{11}+\frac {1}{11}-\frac {1}{13})$

$=-\frac {1}{2}×(1-\frac {1}{13})$

$=-\frac {6}{13}$