Các câu hỏi tương tự
CHo S=\(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2023}{4^{2023}}\). Chứng minh S < \(\dfrac{1}{2}\)
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}}{\dfrac{2022}{1}+\dfrac{2021}{2}+\dfrac{2020}{3}+...+\dfrac{1}{2022}}\)
9 tháng 5 2022 lúc 9:56
Tìm x, biết:
( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2023}\) ) . x = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) + \(\dfrac{2020}{3}\)
+ ... + \(\dfrac{1}{2022}\)
A = \(\dfrac{1}{3}\)-\(\dfrac{2}{^{ }3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+...+\(\dfrac{2023}{3^{2023}}\)-\(\dfrac{2024}{3^{2024}}\) so sánh A với \(\dfrac{3}{16}\)
22 tháng 5 2023 lúc 6:40
( 1 + \(\dfrac{1}{2}\) ) ( 1 + \(\dfrac{1}{3}\) ) ( 1 + \(\dfrac{1}{4}\) ) ... ( 1 + \(\dfrac{1}{2023}\) )
So sánh A và B
A=\(\dfrac{1}{3^1}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{2023}}\)
B=\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{12}\)
22 tháng 3 2023 lúc 11:45
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{2023^2}\) chứng tỏ B<1
giải giúp mình với:
tính tổng của A biết:
A= \(\dfrac{1}{2}\)+ \(\dfrac{1}{3}\)+ \(\dfrac{1}{4}\)+....+ \(\dfrac{1}{2022}\)+\(\dfrac{1}{2023}\)
\(\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{3}\right)\)\(.\left(1+\dfrac{1}{4}\right)...\left(1+\dfrac{1}{2023}\right)\)