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第49页

信息发布者：

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$S = 1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+·s+\frac{1}{2^{2026}}$①，

则$\frac{1}{2}S=\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}+·s+\frac{1}{2^{2027}}$②。

由① - ②，得$\frac{1}{2}S=1-\frac{1}{2^{2027}}，$

所以$S = 2-\frac{1}{2^{2026}}，$即$1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+·s+\frac{1}{2^{2026}}=2-\frac{1}{2^{2026}}。$

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

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

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

$=-\frac {1}{2}×(1-\frac {1}{13})$

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