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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 . \(\dfrac{1}{8.9}\)
a/ Ta có :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...........+\dfrac{1}{n^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.......................
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{n}< 1\)
\(\Leftrightarrow A< 1\)
b/ Ta có :
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+.................+\dfrac{1}{\left(2n\right)^2}\)
\(=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{n^2}\right)\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..................
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+.........+\dfrac{1}{\left(n-1\right)n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+1-\dfrac{1}{n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{2}-\dfrac{1}{4n}< \dfrac{1}{2}\)
\(\Leftrightarrow B< \dfrac{1}{2}\)
\(\)\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(A< 1-\dfrac{1}{n}< 1\)
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2n^2}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)
\(B=\dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)
\(B< \dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{\left(n-1\right)n}\right)\)
Ta có:\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)
\(A=1-\dfrac{1}{2^{100}}< 1\)
Vậy A<1
Câu 1:
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3x}+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}x+\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{1}{3}\)
\(\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{1}{2}\)
\(\Rightarrow x.\dfrac{4}{3}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{3}{8}\)
\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\\ 3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\\ 3A-A=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\right)\\ 2A=3-\dfrac{1}{3^{2014}}\\ A=\left(3-\dfrac{1}{3^{2014}}\right):2\\ A=3:2-\dfrac{1}{3^{2014}}:2\\ A=\dfrac{3}{2}-\dfrac{1}{3^{2014}\cdot2}< \dfrac{3}{2}\)
Vậy \(A< \dfrac{3}{2}\)
Bài 2:
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)
Vậy A < 2
Bài 3:
D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)
\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)
Bài 4:
A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)
\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)
\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
a) Giải
Ta có: \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}+\dfrac{1}{2^{2013}}\)
\(\Rightarrow2S=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{2012}}+\dfrac{2}{2^{2013}}\)
\(2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)
\(\Rightarrow2S-S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2012}}-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=1-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=\dfrac{2^{2013}-1}{2^{2013}}\)
b) Giải
Từ \(A=\dfrac{2011^{2012}+1}{2011^{2013}+1}\)
\(\Rightarrow2011A=\dfrac{2011^{2013}+20111}{2011^{2013}+1}=\dfrac{2011^{2013}+1+2010}{2011^{2013}+1}=1+\dfrac{2010}{2011^{2013}+1}\)
Từ \(B=\dfrac{2011^{2013}+1}{2011^{2014}+1}\)
\(\Rightarrow2011B=\dfrac{2011^{2014}+2011}{2011^{2014}+1}=\dfrac{2011^{2014}+1+2010}{2011^{2014}+1}=1+\dfrac{2010}{2011^{2014}+1}\)
Vì 20112013 + 1 < 20112014 + 1 và 2010 > 0
\(\Rightarrow\dfrac{2010}{2011^{2013}+1}>\dfrac{2010}{2011^{2014}+1}\)
\(\Rightarrow2011A>2011B\)
\(\Rightarrow A>B\)
Vậy A > B.
Lời giải:
$A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2023}{2^{2023}}$
$2A=1+\frac{2}{2}+\frac{3}{2^2}+....+\frac{2023}{2^{2022}}$
$\Rightarrow 2A-A=(1+\frac{2}{2}+\frac{3}{2^2}+....+\frac{2023}{2^{2022}})-(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2023}{2^{2023}})$
$\Rightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2022}}-\frac{2023}{2^{2023}}$
$\Rightarrow A-\frac{2023}{2^{2023}}=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2022}}$
$\Rightarrow 2(A-\frac{2023}{2^{2023}})=2+1+\frac{1}{2}+....+\frac{1}{2^{2021}}$
$\Rightarrow 2(A-\frac{2023}{2^{2023}})-(A-\frac{2023}{2^{2023}})=2-\frac{1}{2^{2022}}$
$\Rightarrow A-\frac{2023}{2^{2023}}=2-\frac{1}{2^{2022}}$
$\Rightarrow A=2-\frac{1}{2^{2022}}+\frac{2023}{2^{2023}}=2+\frac{2021}{2^{2023}}>2$