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8. 计算：$\frac{1}{2026} + \frac{2}{2026} + \frac{3}{2026} + \dots + \frac{4051}{2026} = \_\_\_\_\_\_$
答案
4 051 解析:设 $A=\dfrac{1}{2\ 026}+\dfrac{2}{2\ 026}+\dfrac{3}{2\ 026}+\dots+\dfrac{4\ 051}{2\ 026},B=\dfrac{4\ 051}{2\ 026}+\dfrac{4\ 050}{2\ 026}+\dots+\dfrac{3}{2\ 026}+\dfrac{2}{2\ 026}+\dfrac{1}{2\ 026}$,则 $A=B,A+B=2A=(\dfrac{1}{2\ 026}+\dfrac{4\ 051}{2\ 026})+(\dfrac{2}{2\ 026}+\dfrac{4\ 050}{2\ 026})+\dots+(\dfrac{4\ 050}{2\ 026}+\dfrac{2}{2\ 026})+(\dfrac{4\ 051}{2\ 026}+\dfrac{1}{2\ 026})=\dfrac{4\ 052}{2\ 026}+\dfrac{4\ 052}{2\ 026}+\dots+\dfrac{4\ 052}{2\ 026}=2×4\ 051=8\ 102$,所以 $A=4\ 051$.
9. 计算：$\frac{1}{2}+(\frac{1}{3}+\frac{2}{3})+(\frac{1}{4}+\frac{2}{4}+\frac{3}{4})+(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5})+\dots+(\frac{1}{50}+\frac{2}{50}+\frac{3}{50}+\dots+\frac{49}{50})$
答案
设 $S=\dfrac{1}{2}+(\dfrac{1}{3}+\dfrac{2}{3})+(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4})+(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5})+\dots+(\dfrac{1}{50}+\dfrac{2}{50}+\dots+\dfrac{48}{50}+\dfrac{49}{50})$,则有 $S=\dfrac{1}{2}+(\dfrac{2}{3}+\dfrac{1}{3})+(\dfrac{3}{4}+\dfrac{2}{4}+\dfrac{1}{4})+(\dfrac{4}{5}+\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{1}{5})+\dots+(\dfrac{49}{50}+\dfrac{48}{50}+\dots+\dfrac{2}{50}+\dfrac{1}{50})$,所以 $S+S=\dfrac{1}{2}+(\dfrac{1}{3}+\dfrac{2}{3})+(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4})+(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5})+\dots+(\dfrac{1}{50}+\dfrac{2}{50}+\dots+\dfrac{48}{50}+\dfrac{49}{50})+\dfrac{1}{2}+(\dfrac{2}{3}+\dfrac{1}{3})+(\dfrac{3}{4}+\dfrac{2}{4}+\dfrac{1}{4})+(\dfrac{4}{5}+\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{1}{5})+\dots+(\dfrac{49}{50}+\dfrac{48}{50}+\dots+\dfrac{2}{50}+\dfrac{1}{50})$,即 $2S=1+2+3+\dots+49=1\ 225$,故原式$=\dfrac{1\ 225}{2}$.
一题多解 原式$=\dfrac{1}{2}×1+\dfrac{1}{2}×2+\dfrac{1}{2}×3+\dots+\dfrac{1}{2}×49=\dfrac{1}{2}×(1+2+3+\dots+49)=\dfrac{1\ 225}{2}$.
一题多解 原式$=\dfrac{1}{2}×1+\dfrac{1}{2}×2+\dfrac{1}{2}×3+\dots+\dfrac{1}{2}×49=\dfrac{1}{2}×(1+2+3+\dots+49)=\dfrac{1\ 225}{2}$.
10. 计算:333×999−1 002×332=
3
.答案
3 解析:$333×999=333×(1\ 000-1)=333×1\ 000-333$,$1\ 002×332=(1\ 000+2)×332=332×1\ 000+664$,所以 $333×999-1\ 002×332=333×1\ 000-332×1\ 000-333-664=1\ 000-333-664=3$.
11. 计算: $1+3\frac{1}{6}+5\frac{1}{12}+7\frac{1}{20}+9\frac{1}{30}+11\frac{1}{42}+13\frac{1}{56}+15\frac{1}{72}+17\frac{1}{90}.$
答案
原式$=(1+3+5+7+9+11+13+15+17)+(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90})=81+(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10})=81+(\dfrac{1}{2}-\dfrac{1}{10})=81+\dfrac{2}{5}=81\ \dfrac{2}{5}$.
12. 计算：
(1) $\frac{1}{2} + (\frac{1}{2})^2 + (\frac{1}{2})^3 + (\frac{1}{2})^4 + \dots + (\frac{1}{2})^8$；
(2) $5 + 2 × 5^2 + 3 × 5^3 + 4 × 5^4 + \dots + 8 × 5^8$.(结果可保留指数)
(1) $\frac{1}{2} + (\frac{1}{2})^2 + (\frac{1}{2})^3 + (\frac{1}{2})^4 + \dots + (\frac{1}{2})^8$；
(2) $5 + 2 × 5^2 + 3 × 5^3 + 4 × 5^4 + \dots + 8 × 5^8$.(结果可保留指数)
答案
(1)设 $S=\dfrac{1}{2}+(\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4+\dots+(\dfrac{1}{2})^8$ ①,
则 $2S=1+\dfrac{1}{2}+(\dfrac{1}{2})^2+(\dfrac{1}{2})^3+\dots+(\dfrac{1}{2})^7$ ②,
则②-①,得 $2S-S=S=1-\dfrac{1}{2^8}=\dfrac{255}{256}$.
(2)设 $M=5+2×5^2+3×5^3+4×5^4+\dots+8×5^8$ ①,所以 $5M=1×5^2+2×5^3+3×5^4+\dots+8×5^9$ ②,所以②-①得 $5M-M=(1×5^2+2×5^3+3×5^4+\dots+8×5^9)-(5+2×5^2+3×5^3+4×5^4+\dots+8×5^8)=8×5^9-(5+5^2+5^3+\dots+5^8)$,设 $T=5+5^2+5^3+\dots+5^8$,所以 $5T=5^2+5^3+\dots+5^9$,所以 $5T-T=5^9-5$,所以 $4T=5^9-5$,所以 $T=\dfrac{5^9-5}{4}$,所以 $5M-M=8×5^9-\dfrac{5^9-5}{4}$,所以 $4M=8×5^9-\dfrac{5^9-5}{4}$,所以 $M=\dfrac{8×5^9-\dfrac{5^9-5}{4}}{4}=\dfrac{31×5^9+5}{16}$.
则 $2S=1+\dfrac{1}{2}+(\dfrac{1}{2})^2+(\dfrac{1}{2})^3+\dots+(\dfrac{1}{2})^7$ ②,
则②-①,得 $2S-S=S=1-\dfrac{1}{2^8}=\dfrac{255}{256}$.
(2)设 $M=5+2×5^2+3×5^3+4×5^4+\dots+8×5^8$ ①,所以 $5M=1×5^2+2×5^3+3×5^4+\dots+8×5^9$ ②,所以②-①得 $5M-M=(1×5^2+2×5^3+3×5^4+\dots+8×5^9)-(5+2×5^2+3×5^3+4×5^4+\dots+8×5^8)=8×5^9-(5+5^2+5^3+\dots+5^8)$,设 $T=5+5^2+5^3+\dots+5^8$,所以 $5T=5^2+5^3+\dots+5^9$,所以 $5T-T=5^9-5$,所以 $4T=5^9-5$,所以 $T=\dfrac{5^9-5}{4}$,所以 $5M-M=8×5^9-\dfrac{5^9-5}{4}$,所以 $4M=8×5^9-\dfrac{5^9-5}{4}$,所以 $M=\dfrac{8×5^9-\dfrac{5^9-5}{4}}{4}=\dfrac{31×5^9+5}{16}$.
13. 计算:
(1) $(\frac{1}{6}+\frac{1}{7}+\frac{1}{8})-4×(\frac{1}{2}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8})-5×(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}) = \_\_\_\_\_\_;$
(2) $(\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2026})×(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2025})-(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2026})×(\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2025}) = \_\_\_\_\_\_.$
(1) $(\frac{1}{6}+\frac{1}{7}+\frac{1}{8})-4×(\frac{1}{2}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8})-5×(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}) = \_\_\_\_\_\_;$
(2) $(\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2026})×(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2025})-(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2026})×(\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2025}) = \_\_\_\_\_\_.$
答案
(1)$-1\ \dfrac{4}{9}$ 解析:把 $\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}$ 当成一个整体,则原式$=(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8})-4×[\dfrac{1}{2}-(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8})]-5×[(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8})-\dfrac{1}{9}]=(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8})-2+4×(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8})-5×(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8})+\dfrac{5}{9}=-2+\dfrac{5}{9}=-1\ \dfrac{4}{9}$.
(2)$\dfrac{1}{2\ 026}$ 解析:设 $a=\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{2\ 026},b=\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{2\ 025}$,则原式$=a(b+1)-(a+1)b=ab+a-ab-b=a-b=\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{2\ 026}-(\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{2\ 025})=\dfrac{1}{2\ 026}$.
(2)$\dfrac{1}{2\ 026}$ 解析:设 $a=\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{2\ 026},b=\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{2\ 025}$,则原式$=a(b+1)-(a+1)b=ab+a-ab-b=a-b=\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{2\ 026}-(\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{2\ 025})=\dfrac{1}{2\ 026}$.
