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Đây là dạng toán nâng cao chuyên đề về so sánh phân số, cấu trúc thi chuyên, thi học sinh giỏi, thi violympic. Hôm nay olm sẽ hướng dẫn em cách giải dạng này như sau.
Xét dãy số: 2; 3; 4;...; 2023
Dãy số trên là dãy số cách đều với khoảng cách là: 2 - 1 = 1
Số số hạng của dãy số trên là: (2023 - 2) : 1 + 1 = 2022
Vì \(\dfrac{3}{2^2}\) = \(\dfrac{3}{4}\) < 1 ; \(\dfrac{8}{3^2}\) = \(\dfrac{3^2-1}{3^2}\) < 1;...; \(\dfrac{2023^2-1}{2023^2}\) < 1
Vậy A là tổng của 2022 phân số mã mỗi phân số đều nhỏ hơn 1
⇒ A < 1 x 2022 = 2022 (1)
Mặt khác ta có:
A = \(\dfrac{3}{2^2}\) + \(\dfrac{8}{3^2}\) + \(\dfrac{15}{4^2}\) + \(\dfrac{2023^2-1}{2023^2}\)
A = 1 - \(\dfrac{1}{2^2}\) + 1 - \(\dfrac{1}{3^2}\) + ... + 1 - \(\dfrac{1}{2023^2}\)
A = (1 + 1 + 1+ ...+ 1) - (\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{2023^2}\))
A = 2022 - (\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + .... + \(\dfrac{1}{2023^2}\))
Đặt B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + .... + \(\dfrac{1}{2023^2}\)
\(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)
\(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{4^2}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)
............................
\(\dfrac{1}{2023^2}\)< \(\dfrac{1}{2022.2023}\) = \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\)
Cộng vế với vế ta có:
B < 1 - \(\dfrac{1}{2023}\)
⇒ - B > -1 + \(\dfrac{1}{2023}\)
⇒ A = 2022 - B > 2022 - 1 + \(\dfrac{1}{2023}\) = 2021 + \(\dfrac{1}{2023}\) ⇒ A > 2021 (2)
Kết hợp (1) và (2) ta có:
2021 < A < 2022
Vậy A không phải là số tự nhiên (đpcm)
A = 3. \(\dfrac{1}{1.2}\) - 5. \(\dfrac{1}{2.3}\) + 7. \(\dfrac{1}{3.4}\) + ... + 15. \(\dfrac{1}{7.8}\) -17 . \(\dfrac{1}{8.9}\)
1/2^2=4
1/3^2<1/2.3
.................
1/100^2<1/99.100
A<1/4+1/2.3+...+1/99.100
A<1/4+1/2-1/100
A<1/4<3/4
Vậy A<3/4(dpcm).CHÚC BẠN HỌC TỐT!
Ta có: \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+...+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+...+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)
\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\)
⇒5A=15+252+...+11511⇒5A=15+252+...+11511
⇒5A−A=15+152+...+1511−11512⇒5A−A=15+152+...+1511−11512
⇒4A=15+152+...+1511−11512⇒4A=15+152+...+1511−11512
⇒20A=1+15+...+1510−11511⇒20A=1+15+...+1510−11511
⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)
⇒16A=1−12511+11512<1⇒16A=1−12511+11512<1
⇒A<116⇒A<116
a) M =1+3+32+33+......+3118+3119
M = ( 1+3+32 ) +...+ ( 3117 + 3118+3119 )
M = 1. ( 1+3+32 ) + ... + 3117 . ( 3117 + 3118+3119 )
M = ( 1+3+32 ) .( 1 + ... + 3117 )
M = 13 . ( 1 + ... + 3117 ) \(⋮\) 13 (đpcm )
b) Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{2009^2}< \dfrac{1}{2008.2009}\)
\(\dfrac{1}{2010^2}< \dfrac{1}{2009.2010}\)
=> \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\) (1)
Biến đổi vế trái:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2010}\)
= \(1-\dfrac{1}{2010}\)
= \(\dfrac{2009}{2010}< 1\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < 1 hay:
N < 1
Đặt \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)
Ta có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2008^2}< \)\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2007\cdot2008}\left(1\right)\)
Lại có: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)
\(=1-\dfrac{1}{2008}< 1\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\) ta có \(A< B< 1\Rightarrow A< 1\)
A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{2008^2}\)
A<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2007.2008}\)
A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)
A<\(1-\dfrac{1}{2008}\)
A<\(\dfrac{2007}{2008}< 1\)
=> A<1
Vậy A<1
Kiyoko Vũ
a, xét từng đoạn 1 , 1/2 ,1/2^3 ,1/2^4 ,1/2^5 ,1/2^6
ta có
1 = 1
1/2 + 1/3 < 1/2 + 1/2 = 1
1/4 + 1/5 + .. + 1/7 < 1/4 +..+ 1/4 = 4/4 = 1
1/8 + 1/9 + .. + 1/15 < 1/8 + .. + 1/8 = 8/8 = 1
tương tự
1/16 +1/17 + .. + 1/31 < 1
1/32 + 1/33 + .. + 1/63 < 1
=> cộng lại => A < 6
b, Câu hỏi của trịnh quỳnh trang - Toán lớp 6 - Học toán với OnlineMath
Ta có:
A=\(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
A<\(1+\dfrac{1}{2.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
A<\(1+\dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
A<\(\dfrac{5}{4}\)+\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{99}-\dfrac{1}{100}\)
A<\(\dfrac{5}{4}+\dfrac{1}{2}-\dfrac{1}{100}\)
A<\(\dfrac{5}{4}+\dfrac{49}{100}\)
A<\(\dfrac{174}{100}\)<\(\dfrac{7}{4}\)
=>A<\(\dfrac{7}{4}\)
Tick giùm mink nha :D
Đáp án nè:
Đặt A=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{99}}\)
3A=\(\dfrac{1}{1}-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
3A+A=\(\left(\dfrac{1}{1}-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
4A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)
4A bé hơn(sorry tớ không thấy dấu bé hơn)\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
Đặt B=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
3B=\(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)
4B=\(3-\dfrac{1}{3^{99}}\) bé hơn 3 \(\Rightarrow\)B bé hơn \(\dfrac{3}{4}\)
\(\Rightarrow\) 4A bé hơn\(\dfrac{3}{4}\Rightarrow\)A bé hơn \(\dfrac{3}{16}\)
Tick cho mình nha , ngồi đánh máy tính mỏi cả mắt lun
Chúc học tốt
\(A=\dfrac{2}{3}+\dfrac{2}{3^2}+\dfrac{2}{3^3}+....+\dfrac{2}{3^{2023}}\)
\(3A=2+\dfrac{2}{3}+\dfrac{2}{3^2}+....+\dfrac{2}{3^{2022}}\)
\(3A-A=\left(2+\dfrac{2}{3}+\dfrac{2}{3^2}+...+\dfrac{2}{3^{2022}}\right)-\left(\dfrac{2}{3}+\dfrac{2}{3^2}+....+\dfrac{2}{3^{2023}}\right)\)
\(2A=2-\dfrac{2}{3^{2023}}\)
\(A=\left(2-\dfrac{2}{3^{2023}}\right)\times\dfrac{1}{2}\)
\(A=2\times\dfrac{1}{2}-\dfrac{2}{3^{2023}}\times\dfrac{1}{2}\)
\(A=1-\dfrac{1}{3^{2023}}\)
=> \(A< 1\left(đpcm\right)\)