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1. 探索规律，观察下面算式，解答问题.
$1+3=4=2^2$；$1+3+5=9=3^2$；$1+3+5+7=16=4^2$；$1+3+5+7+9=25=5^2$；….
(1)请猜想$1+3+5+7+9+\dots+19=$
(2)请猜想$1+3+5+7+9+\dots+(2n-1)=$
(3)试计算:$1\ 021+1\ 023+\dots+2\ 025+2\ 027$.
$1+3=4=2^2$；$1+3+5=9=3^2$；$1+3+5+7=16=4^2$；$1+3+5+7+9=25=5^2$；….
(1)请猜想$1+3+5+7+9+\dots+19=$
100
；(2)请猜想$1+3+5+7+9+\dots+(2n-1)=$
$n^2$
；($n$是整数且$n＞1$)(3)试计算:$1\ 021+1\ 023+\dots+2\ 025+2\ 027$.
答案
(1) 100
(2) $n^2$
(3) $1\ 021+1\ 023+\dots+2\ 025+2\ 027$
$=(1+3+5+\dots+2\ 025+2\ 027)-(1+3+5+\dots+1\ 017+1\ 019)$
$=(\dfrac{1+2\ 027}{2})^2-(\dfrac{1+1\ 019}{2})^2$
$=1\ 014^2-510^2$
$=768\ 096$
(2) $n^2$
(3) $1\ 021+1\ 023+\dots+2\ 025+2\ 027$
$=(1+3+5+\dots+2\ 025+2\ 027)-(1+3+5+\dots+1\ 017+1\ 019)$
$=(\dfrac{1+2\ 027}{2})^2-(\dfrac{1+1\ 019}{2})^2$
$=1\ 014^2-510^2$
$=768\ 096$
2. [2025·烟台模拟]阅读材料：
$\frac{1}{1×2}=1-\frac{1}{2},\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3};$
$\frac{1}{1×3}=\frac{1}{2}×(1-\frac{1}{3}),\frac{1}{3×5}=\frac{1}{2}×(\frac{1}{3}-\frac{1}{5});$
$\frac{1}{1×5}=\frac{1}{4}×(1-\frac{1}{5}),\frac{1}{5×9}=\frac{1}{4}×(\frac{1}{5}-\frac{1}{9}),$
$\frac{1}{9×13}=\frac{1}{4}×(\frac{1}{9}-\frac{1}{13}).$
请根据以上各式解答下列问题：
(1)仿照阅读材料，将下列算式变形：
$\frac{1}{5×7}=\_\_\_\_\_\_,\frac{1}{13×17}=\_\_\_\_\_\_,$
$\frac{1}{2021×2026}=\_\_\_\_\_\_;$
(2)计算：$\frac{1}{1×6}+\frac{1}{6×11}+\frac{1}{11×16}+…+\frac{1}{2021×2026};$
(3)计算：$\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+…+\frac{1}{210}.$
$\frac{1}{1×2}=1-\frac{1}{2},\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3};$
$\frac{1}{1×3}=\frac{1}{2}×(1-\frac{1}{3}),\frac{1}{3×5}=\frac{1}{2}×(\frac{1}{3}-\frac{1}{5});$
$\frac{1}{1×5}=\frac{1}{4}×(1-\frac{1}{5}),\frac{1}{5×9}=\frac{1}{4}×(\frac{1}{5}-\frac{1}{9}),$
$\frac{1}{9×13}=\frac{1}{4}×(\frac{1}{9}-\frac{1}{13}).$
请根据以上各式解答下列问题：
(1)仿照阅读材料，将下列算式变形：
$\frac{1}{5×7}=\_\_\_\_\_\_,\frac{1}{13×17}=\_\_\_\_\_\_,$
$\frac{1}{2021×2026}=\_\_\_\_\_\_;$
(2)计算：$\frac{1}{1×6}+\frac{1}{6×11}+\frac{1}{11×16}+…+\frac{1}{2021×2026};$
(3)计算：$\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+…+\frac{1}{210}.$
答案
(1) $\frac{1}{2}×(\frac{1}{5}-\frac{1}{7})$；$\frac{1}{4}×(\frac{1}{13}-\frac{1}{17})$；$\frac{1}{5}×(\frac{1}{2\ 021}-\frac{1}{2\ 026})$
(2) $\frac{1}{1×6}+\frac{1}{6×11}+\frac{1}{11×16}+\dots+\frac{1}{2021×2026}$
$=\frac{1}{5}×(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\dots+\frac{1}{2\ 021}-\frac{1}{2\ 026})$
$=\frac{1}{5}×(1-\frac{1}{2\ 026})$
$=\frac{1}{5}×\frac{2\ 025}{2\ 026}$
$=\frac{405}{2\ 026}$
(3) $\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\dots+\frac{1}{210}$
$=\frac{1}{3×4}+\frac{1}{4×5}+\frac{1}{5×6}+\dots+\frac{1}{14×15}$
$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\dots+\frac{1}{14}-\frac{1}{15}$
$=\frac{1}{3}-\frac{1}{15}$
$=\frac{4}{15}$
(2) $\frac{1}{1×6}+\frac{1}{6×11}+\frac{1}{11×16}+\dots+\frac{1}{2021×2026}$
$=\frac{1}{5}×(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\dots+\frac{1}{2\ 021}-\frac{1}{2\ 026})$
$=\frac{1}{5}×(1-\frac{1}{2\ 026})$
$=\frac{1}{5}×\frac{2\ 025}{2\ 026}$
$=\frac{405}{2\ 026}$
(3) $\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\dots+\frac{1}{210}$
$=\frac{1}{3×4}+\frac{1}{4×5}+\frac{1}{5×6}+\dots+\frac{1}{14×15}$
$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\dots+\frac{1}{14}-\frac{1}{15}$
$=\frac{1}{3}-\frac{1}{15}$
$=\frac{4}{15}$
3. 先观察下列等式，再回答问题：
①$1×3+1=2^2$；②$2×4+1=3^2$；③$3×5+1=4^2$；④$4×6+1=5^2$；….
(1)写出第5个等式：
(2)写出第$n$($n$为正整数)个等式：
(3)请利用上述规律计算：$(1+\dfrac{1}{1×3})×(1+\dfrac{1}{2×4})×(1+\dfrac{1}{3×5})×\dots×(1+\dfrac{1}{100×102})$.
①$1×3+1=2^2$；②$2×4+1=3^2$；③$3×5+1=4^2$；④$4×6+1=5^2$；….
(1)写出第5个等式：
$5×7+1=6^2$
；(2)写出第$n$($n$为正整数)个等式：
$n(n+2)+1=(n+1)^2$
；(3)请利用上述规律计算：$(1+\dfrac{1}{1×3})×(1+\dfrac{1}{2×4})×(1+\dfrac{1}{3×5})×\dots×(1+\dfrac{1}{100×102})$.
答案
(1) $5×7+1=6^2$
(2) $n(n+2)+1=(n+1)^2$
(3) 原式$=\dfrac{1×3+1}{1×3}×\dfrac{2×4+1}{2×4}×\dfrac{3×5+1}{3×5}×\dots×\dfrac{100×102+1}{100×102}$
$=\dfrac{2^2}{1×3}×\dfrac{3^2}{2×4}×\dfrac{4^2}{3×5}×\dots×\dfrac{101^2}{100×102}$
$=2×\dfrac{2}{3}×\dfrac{3}{2}×\dfrac{3}{4}×\dfrac{4}{3}×\dfrac{4}{5}×\dots×\dfrac{101}{100}×\dfrac{101}{102}$
$=2×\dfrac{101}{102}$
$=\dfrac{101}{51}$
(2) $n(n+2)+1=(n+1)^2$
(3) 原式$=\dfrac{1×3+1}{1×3}×\dfrac{2×4+1}{2×4}×\dfrac{3×5+1}{3×5}×\dots×\dfrac{100×102+1}{100×102}$
$=\dfrac{2^2}{1×3}×\dfrac{3^2}{2×4}×\dfrac{4^2}{3×5}×\dots×\dfrac{101^2}{100×102}$
$=2×\dfrac{2}{3}×\dfrac{3}{2}×\dfrac{3}{4}×\dfrac{4}{3}×\dfrac{4}{5}×\dots×\dfrac{101}{100}×\dfrac{101}{102}$
$=2×\dfrac{101}{102}$
$=\dfrac{101}{51}$
4.阅读材料:求 $5+5^2+5^3+5^4+5^5+5^6+5^7+5^8$ 的值.
解:设 $S=5+5^2+5^3+5^4+5^5+5^6+5^7+5^8$,①
则 $5S=5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9$.②
②$-$①,得 $5S-S=(5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9)-(5+5^2+5^3+5^4+5^5+5^6+5^7+5^8)$,
所以 $4S=5^9-5$,即 $S=\frac{5^9 -5}{4}$.
所以 $5+5^2+5^3+5^4+5^5+5^6+5^7+5^8=\frac{5^9 -5}{4}$.
以上方法我们称为“错位相减法”.
请利用上述材料,解决下列问题:
(1)计算:$\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+(\frac{1}{2})^4+\dots+(\frac{1}{2})^8$;
(仿照材料写出求解过程)
(2)化简:$M=5+2×5^2+3×5^3+4×5^4+\dots+8×5^8$.
解:设 $S=5+5^2+5^3+5^4+5^5+5^6+5^7+5^8$,①
则 $5S=5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9$.②
②$-$①,得 $5S-S=(5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9)-(5+5^2+5^3+5^4+5^5+5^6+5^7+5^8)$,
所以 $4S=5^9-5$,即 $S=\frac{5^9 -5}{4}$.
所以 $5+5^2+5^3+5^4+5^5+5^6+5^7+5^8=\frac{5^9 -5}{4}$.
以上方法我们称为“错位相减法”.
请利用上述材料,解决下列问题:
(1)计算:$\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+(\frac{1}{2})^4+\dots+(\frac{1}{2})^8$;
(仿照材料写出求解过程)
(2)化简:$M=5+2×5^2+3×5^3+4×5^4+\dots+8×5^8$.
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 $2S=1+\dfrac{1}{2}+(\dfrac{1}{2})^2+\dots+(\dfrac{1}{2})^7$,②
②$-$①,得 $2S-S=S=1-(\dfrac{1}{2})^8$.
所以 $\dfrac{1}{2}+(\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4+\dots+(\dfrac{1}{2})^8=1-(\dfrac{1}{2})^8$.
(2) 因为 $M=5+2×5^2+3×5^3+4×5^4+\dots+8×5^8$,①
所以 $5M=1×5^2+2×5^3+3×5^4+4×5^5+\dots+8×5^9$,②
②$-$①,得 $5M-M=4M=(1×5^2+2×5^3+3×5^4+4×5^5+\dots+8×5^9)-(5+2×5^2+3×5^3+4×5^4+\dots+8×5^8)=8×5^9-(5+5^2+5^3+5^4+\dots+5^8)$,
设 $T=5+5^2+5^3+5^4+\dots+5^8$,
所以 $5T=5^2+5^3+5^4+5^5+\dots+5^9$.
所以 $5T-T=5^9-5$,
即 $4T=5^9-5$.
所以 $T=\dfrac{5^9-5}{4}$.所以 $4M=8×5^9-\dfrac{5^9-5}{4}$.
所以 $M=\dfrac{8×5^9}{4}-\dfrac{5^9-5}{16}=\dfrac{31×5^9+5}{16}$.
则 $2S=1+\dfrac{1}{2}+(\dfrac{1}{2})^2+\dots+(\dfrac{1}{2})^7$,②
②$-$①,得 $2S-S=S=1-(\dfrac{1}{2})^8$.
所以 $\dfrac{1}{2}+(\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4+\dots+(\dfrac{1}{2})^8=1-(\dfrac{1}{2})^8$.
(2) 因为 $M=5+2×5^2+3×5^3+4×5^4+\dots+8×5^8$,①
所以 $5M=1×5^2+2×5^3+3×5^4+4×5^5+\dots+8×5^9$,②
②$-$①,得 $5M-M=4M=(1×5^2+2×5^3+3×5^4+4×5^5+\dots+8×5^9)-(5+2×5^2+3×5^3+4×5^4+\dots+8×5^8)=8×5^9-(5+5^2+5^3+5^4+\dots+5^8)$,
设 $T=5+5^2+5^3+5^4+\dots+5^8$,
所以 $5T=5^2+5^3+5^4+5^5+\dots+5^9$.
所以 $5T-T=5^9-5$,
即 $4T=5^9-5$.
所以 $T=\dfrac{5^9-5}{4}$.所以 $4M=8×5^9-\dfrac{5^9-5}{4}$.
所以 $M=\dfrac{8×5^9}{4}-\dfrac{5^9-5}{16}=\dfrac{31×5^9+5}{16}$.
