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Lời giải:
Ta có:
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\)
\(S> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}\)
\(\Leftrightarrow S> \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2016-2015}{2015.2016}\)
\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{2016}=\frac{1007}{2016}\)
--------------------------
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}\)
\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2014}{2015}\)
\(\Leftrightarrow S< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2015-2014}{2014.2015}\)
\(\Leftrightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2014}-\frac{1}{2015}\)
\(\Leftrightarrow S< 1-\frac{1}{2015}=\frac{2014}{2015}\)
Vậy ta có đpcm.
\(S=2S-S\)
\(=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^{2022}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^{2022}}\right)\)
\(=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^{2022}}\right)\)
\(=2-\dfrac{1}{2^{2022}}< 2\left(đpcm\right)\)
Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
= 1/2-1/3+1/3-1/4+...+1/10-1/11
= 1/2 - 1/11 = 9/22 (đpcm)
1/2^2+1/3^2+...+1/50^2<1/1*2+1/2*3*+...+1/49*50
=1/1-1/2+1/2-1/3+...+1/49-1/50<1
=>S<1+1=2
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b: 
S=\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2018.2018}\)
Ta có: \(\dfrac{1}{5.5}< \dfrac{1}{4.5};\dfrac{1}{6.6}< \dfrac{1}{5.6};\dfrac{1}{7.7}< \dfrac{1}{6.7};...;\dfrac{1}{2018.2018}< \dfrac{1}{2017.2018}\)
\(\Rightarrow\) S<\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{2017.2018}\)
S<\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)
S< \(\dfrac{1}{4}-\dfrac{1}{2018}< \dfrac{1}{4}\)
\(\Rightarrow\)S<\(\dfrac{1}{4}\)
Học tốt nha
\(S=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A< 1-\dfrac{1}{50}\Rightarrow A< 1\)
Ta có \(S=\dfrac{1}{2^2}\left(1+A\right)\)
Ta có
\(A< 1\Rightarrow1+A< 2\Rightarrow S< \dfrac{1}{2^2}.2=\dfrac{1}{2}\)
đề có bị sai k bn ơi mik thấy cứ sao sao í
ừ \(\dfrac{1}{2024^1}\) sai không bạn?