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4. 计算：
（1）$-1 - 2 - 3 - 4 - \dots - 199 - 200$；
（2）$1 + 5 + 5^2 + 5^3 + \dots + 5^{217} + 5^{218}$。
（1）$-1 - 2 - 3 - 4 - \dots - 199 - 200$；
（2）$1 + 5 + 5^2 + 5^3 + \dots + 5^{217} + 5^{218}$。
答案
（1）设$S=-1-2-3-4-\dots-199-200$①,则$S=-200-199-198-197-\dots-2-1$②. 由①$+$②,得$2S=-201×200$,即$2S=-40\ 200$,所以$S=-20\ 100$,即$-1-2-3-4-\dots-199-200=-20\ 100$
（2）令$S=1+5+5^2+5^3+\dots+5^{217}+5^{218}$①,则$5S=5+5^2+5^3+5^4+\dots+5^{218}+5^{219}$②. 由②$-$①,得$5S-S=5^{219}-1$,所以$S=\dfrac{5^{219}-1}{4}$,即$1+5+5^2+5^3+\dots+5^{217}+5^{218}=\dfrac{5^{219}-1}{4}$
（2）令$S=1+5+5^2+5^3+\dots+5^{217}+5^{218}$①,则$5S=5+5^2+5^3+5^4+\dots+5^{218}+5^{219}$②. 由②$-$①,得$5S-S=5^{219}-1$,所以$S=\dfrac{5^{219}-1}{4}$,即$1+5+5^2+5^3+\dots+5^{217}+5^{218}=\dfrac{5^{219}-1}{4}$
5. 计算: $1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots+\frac{1}{2^{2026}}.$
答案
设$S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dots+\dfrac{1}{2^{2026}}$①,则$\dfrac{1}{2}S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dots+\dfrac{1}{2^{2027}}$②. 由①$-$②,得$\dfrac{1}{2}S=1-\dfrac{1}{2^{2027}}$,所以$S=2-\dfrac{1}{2^{2026}}$,即$1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dots+\dfrac{1}{2^{2026}}=2-\dfrac{1}{2^{2026}}$
6. 阅读材料：
在计算$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\dots+\frac{1}{420}$时，直接计算很繁琐，我们可以采用“裂项——消项”法简化运算。
$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\dots+\frac{1}{420}=\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{4×5}+\dots+\frac{1}{20×21}=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\dots+(\frac{1}{20}-\frac{1}{21})=1-\frac{1}{21}=\frac{20}{21}$。
方法应用：
试用“裂项——消项”法解下面各题：
(1) $\frac{1}{3×7}+\frac{1}{7×11}+\frac{1}{11×15}+\dots+\frac{1}{55×59}$；
(2) $-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}$。
在计算$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\dots+\frac{1}{420}$时，直接计算很繁琐，我们可以采用“裂项——消项”法简化运算。
$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\dots+\frac{1}{420}=\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{4×5}+\dots+\frac{1}{20×21}=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\dots+(\frac{1}{20}-\frac{1}{21})=1-\frac{1}{21}=\frac{20}{21}$。
方法应用：
试用“裂项——消项”法解下面各题：
(1) $\frac{1}{3×7}+\frac{1}{7×11}+\frac{1}{11×15}+\dots+\frac{1}{55×59}$；
(2) $-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}$。
答案
(1) 原式$=\dfrac{1}{4}×(\dfrac{1}{3}-\dfrac{1}{7})+\dfrac{1}{4}×(\dfrac{1}{7}-\dfrac{1}{11})+\dfrac{1}{4}×(\dfrac{1}{11}-\dfrac{1}{15})+\dots+\dfrac{1}{4}×(\dfrac{1}{55}-\dfrac{1}{59})=\dfrac{1}{4}×(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dots+\dfrac{1}{55}-\dfrac{1}{59})=\dfrac{1}{4}×(\dfrac{1}{3}-\dfrac{1}{59})=\dfrac{14}{177}$
(2) 原式$=-\dfrac{1}{1×3}-\dfrac{1}{3×5}-\dfrac{1}{5×7}-\dfrac{1}{7×9}-\dfrac{1}{9×11}-\dfrac{1}{11×13}=-\dfrac{1}{2}×(1-\dfrac{1}{3})-\dfrac{1}{2}×(\dfrac{1}{3}-\dfrac{1}{5})-\dfrac{1}{2}×(\dfrac{1}{5}-\dfrac{1}{7})-\dfrac{1}{2}×(\dfrac{1}{7}-\dfrac{1}{9})-\dfrac{1}{2}×(\dfrac{1}{9}-\dfrac{1}{11})-\dfrac{1}{2}×(\dfrac{1}{11}-\dfrac{1}{13})=-\dfrac{1}{2}×(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13})=-\dfrac{1}{2}×(1-\dfrac{1}{13})=-\dfrac{6}{13}$
(2) 原式$=-\dfrac{1}{1×3}-\dfrac{1}{3×5}-\dfrac{1}{5×7}-\dfrac{1}{7×9}-\dfrac{1}{9×11}-\dfrac{1}{11×13}=-\dfrac{1}{2}×(1-\dfrac{1}{3})-\dfrac{1}{2}×(\dfrac{1}{3}-\dfrac{1}{5})-\dfrac{1}{2}×(\dfrac{1}{5}-\dfrac{1}{7})-\dfrac{1}{2}×(\dfrac{1}{7}-\dfrac{1}{9})-\dfrac{1}{2}×(\dfrac{1}{9}-\dfrac{1}{11})-\dfrac{1}{2}×(\dfrac{1}{11}-\dfrac{1}{13})=-\dfrac{1}{2}×(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13})=-\dfrac{1}{2}×(1-\dfrac{1}{13})=-\dfrac{6}{13}$
