《2025年综合应用创新题典中点六年级数学上册鲁教版五四制》

1. 探索规律,观察下面算式,解答问题.
$1 + 3 = 4 = 2^2$；$1 + 3 + 5 = 9 = 3^2$；$1 + 3 + 5 + 7 = 16 = 4^2$；$1 + 3 + 5 + 7 + 9 = 25 = 5^2$；…$$.
(1)请猜想$1 + 3 + 5 + 7 + 9 + … + 19 = $100；
(2)请猜想$1 + 3 + 5 + 7 + 9 + … + (2n - 1) = $$n^{2}$；($n是整数且n ＞ 1$)
(3)试计算：$1021 + 1023 + … + 2025 + 2027$.
$1021 + 1023 + … + 2025 + 2027=(1 + 3 + 5 + … + 2025 + 2027)-(1 + 3 + 5 + … + 1017 + 1019)=(\frac{1 + 2027}{2})^{2}-(\frac{1 + 1019}{2})^{2}=1014^{2}-510^{2}=768096.$

答案：
(1)100
(2)$n^{2}$
(3)$1021 + 1023 + … + 2025 + 2027=(1 + 3 + 5 + … + 2025 + 2027)-(1 + 3 + 5 + … + 1017 + 1019)=(\frac{1 + 2027}{2})^{2}-(\frac{1 + 1019}{2})^{2}=1014^{2}-510^{2}=768096.$

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} = $$\frac{1}{2}×(\frac{1}{5}-\frac{1}{7})$,$\frac{1}{13×17} = $$\frac{1}{4}×(\frac{1}{13}-\frac{1}{17})$,$\frac{1}{2021×2026} = $$\frac{1}{5}×(\frac{1}{2021}-\frac{1}{2026})$；
(2)计算：$\frac{1}{1×6} + \frac{1}{6×11} + \frac{1}{11×16} + … + \frac{1}{2021×2026}$；
$\frac{1}{1×6}+\frac{1}{6×11}+\frac{1}{11×16}+…+\frac{1}{2021×2026}=\frac{1}{5}×(1 - \frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+…+\frac{1}{2021}-\frac{1}{2026})=\frac{1}{5}×(1 - \frac{1}{2026})=\frac{1}{5}×\frac{2025}{2026}=\frac{405}{2026}.$
(3)计算：$\frac{1}{12} + \frac{1}{20} + \frac{1}{30} + … + \frac{1}{210}$.
$\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+…+\frac{1}{210}=\frac{1}{3×4}+\frac{1}{4×5}+\frac{1}{5×6}+…+\frac{1}{14×15}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+…+\frac{1}{14}-\frac{1}{15}=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}.$

答案：
(1)$\frac{1}{2}×(\frac{1}{5}-\frac{1}{7})$,$\frac{1}{4}×(\frac{1}{13}-\frac{1}{17})$,$\frac{1}{5}×(\frac{1}{2021}-\frac{1}{2026})$
(2)$\frac{1}{1×6}+\frac{1}{6×11}+\frac{1}{11×16}+…+\frac{1}{2021×2026}=\frac{1}{5}×(1 - \frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+…+\frac{1}{2021}-\frac{1}{2026})=\frac{1}{5}×(1 - \frac{1}{2026})=\frac{1}{5}×\frac{2025}{2026}=\frac{405}{2026}.$
(3)$\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+…+\frac{1}{210}=\frac{1}{3×4}+\frac{1}{4×5}+\frac{1}{5×6}+…+\frac{1}{14×15}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+…+\frac{1}{14}-\frac{1}{15}=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}.$

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：$5×7 + 1 = 6^{2}$；
(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：$(1 + \frac{1}{1×3})×(1 + \frac{1}{2×4})×(1 + \frac{1}{3×5})×…×(1 + \frac{1}{100×102})$.
原式$=\frac{1×3 + 1}{1×3}×\frac{2×4 + 1}{2×4}×\frac{3×5 + 1}{3×5}×…×\frac{100×102 + 1}{100×102}=\frac{2^{2}}{1×3}×\frac{3^{2}}{2×4}×\frac{4^{2}}{3×5}×…×\frac{101^{2}}{100×102}=2×\frac{2}{3}×\frac{3}{2}×\frac{3}{4}×\frac{4}{3}×\frac{4}{5}×…×\frac{101}{100}×\frac{101}{102}=2×\frac{101}{102}=\frac{101}{51}.$

答案：
(1)$5×7 + 1 = 6^{2}$
(2)$n(n + 2)+1=(n + 1)^{2}$
(3)原式$=\frac{1×3 + 1}{1×3}×\frac{2×4 + 1}{2×4}×\frac{3×5 + 1}{3×5}×…×\frac{100×102 + 1}{100×102}=\frac{2^{2}}{1×3}×\frac{3^{2}}{2×4}×\frac{4^{2}}{3×5}×…×\frac{101^{2}}{100×102}=2×\frac{2}{3}×\frac{3}{2}×\frac{3}{4}×\frac{4}{3}×\frac{4}{5}×…×\frac{101}{100}×\frac{101}{102}=2×\frac{101}{102}=\frac{101}{51}.$

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：$\frac{1}{2} + (\frac{1}{2})^2 + (\frac{1}{2})^3 + (\frac{1}{2})^4 + … + (\frac{1}{2})^8$；(仿照材料写出求解过程)
(2)化简：$M = 5 + 2×5^2 + 3×5^3 + 4×5^4 + … + 8×5^8$.

答案：
(1)设$S=\frac{1}{2}+(\frac{1}{2})^{2}+(\frac{1}{2})^{3}+(\frac{1}{2})^{4}+…+(\frac{1}{2})^{8}$,① 则$2S=1+\frac{1}{2}+(\frac{1}{2})^{2}+…+(\frac{1}{2})^{7}$,② ② - ①,得$2S - S = S = 1 - (\frac{1}{2})^{8}$. 所以$\frac{1}{2}+(\frac{1}{2})^{2}+(\frac{1}{2})^{3}+(\frac{1}{2})^{4}+…+(\frac{1}{2})^{8}=1 - (\frac{1}{2})^{8}.$
(2)因为$M = 5 + 2×5^{2}+3×5^{3}+4×5^{4}+…+8×5^{8}$,① 所以$5M = 1×5^{2}+2×5^{3}+3×5^{4}+4×5^{5}+…+8×5^{9}$,② ② - ①,得$5M - M = 4M=(1×5^{2}+2×5^{3}+3×5^{4}+4×5^{5}+…+8×5^{9})-(5 + 2×5^{2}+3×5^{3}+4×5^{4}+…+8×5^{8})=8×5^{9}-(5 + 5^{2}+5^{3}+5^{4}+…+5^{8})$ 设$T = 5 + 5^{2}+5^{3}+5^{4}+…+5^{8}$, 所以$5T = 5^{2}+5^{3}+5^{4}+5^{5}+…+5^{9}.$ 所以$5T - T = 5^{9}-5$, 即$4T = 5^{9}-5$, 所以$T=\frac{5^{9}-5}{4}$. 所以$4M = 8×5^{9}-\frac{5^{9}-5}{4}$. 所以$M=\frac{8×5^{9}}{4}-\frac{5^{9}-5}{16}=\frac{31×5^{9}+5}{16}.$