4A=1-1/2^2+1/2^4-...+1/2^2018-1/2^2020
=>5A=1-1/2^2022
=>A=1/5-1/5*2^2022<1/5=0,2
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A=1-\(\dfrac{1}{2^2}\)-\(\dfrac{1}{3^2}\)-...-\(\dfrac{1}{2022^2}\) Chứng minh A>\(\dfrac{1}{2022}\)
chứng minh :\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+.............+\dfrac{1}{2020^2}< \dfrac{1}{4}\)
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}}{\dfrac{2022}{1}+\dfrac{2021}{2}+\dfrac{2020}{3}+...+\dfrac{1}{2022}}\)
cho A=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2022}\)
B=\(\dfrac{2021}{1}+\dfrac{2020}{2}+\dfrac{2019}{3}+...+\dfrac{1}{2021}\)
tính tỉ số \(\dfrac{B}{A}\)
9 tháng 5 2022 lúc 9:56
Tìm x, biết:
( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2023}\) ) . x = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) + \(\dfrac{2020}{3}\)
+ ... + \(\dfrac{1}{2022}\)
3 tháng 5 2022 lúc 12:31
a) tìm x :\(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+....+\dfrac{2}{31.344}x=10\)
b)so sánh hai phân số sau : A=\(\dfrac{6^{2020}+1}{6^{2021}+1}\)và B=\(\dfrac{6^{\text{2021}}+1}{\text{6}^{\text{2022}}+1}\)
ét o ét giúp với ạ
17 tháng 3 2022 lúc 20:20
Cho A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ..... + \(\dfrac{1}{2022^2}\) Chứng tỏ rằng A < 1
cho A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.....+\(\dfrac{1}{2^{2020}}\)+\(\dfrac{1}{2^{2021}}\). Chứng tỏ rằng A<\(\dfrac{1}{2}\)
Giúp vs ạ cần gấp
so sánh 2 phân số:
A=\(\dfrac{6^{2020}+1}{6^{2021}+1}\) với B=\(\dfrac{6^{2021}+1}{6^{2022}+1}\)