Tính tổng:
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2019}\)
K
Khách
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Những câu hỏi liên quan
9 tháng 10 2019
Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)
\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)
\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)
29 tháng 3 2020
Đặt \(A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021\)
DT
17 tháng 7 2019
1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)
\(\Rightarrow1+2019^2=2020^2-2.2019\)
\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)
\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)
\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)
\(=2020\)
Vậy M=2020.
2) Xét : \(k\in N;k\ge2\)ta có:
\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)
\(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)
\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)
\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)
Cho \(k=3,4,...,2020.\)Ta có:
\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)
\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)
Vậy \(N=2018\frac{1009}{2020}.\)
11 tháng 7 2019
\(S=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2019^2-\left(2019^2-2\right)}\)
\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)
\(S=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2019^2}-\sqrt{2019^2-2}\right)\)
\(S=\frac{1}{2}\left(-1+\sqrt{2019^2}\right)\)
\(S=\frac{\left(2019-1\right)}{2}=1009\)
11 tháng 7 2019
\(S=\frac{1-\sqrt{3}}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{2019-\sqrt{2019^2-2}}{2019^2-2019^2-2}.\)
\(S=\frac{1-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+...+\frac{2019-\sqrt{2019^2-2}}{-2}.\)
\(-2S=1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}...+2019-\sqrt{2019^2-2}\)
\(-2S=1-\sqrt{2019^2-2}\Rightarrow S=\frac{\sqrt{2019^2-2}-1}{2}\)
AH
Akai Haruma
Giáo viên
23 tháng 2 2020
Lời giải:
$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+......+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$
$4D=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$
Trừ theo vế:
\(3D=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow 12D=4+1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2017}}-\frac{2019}{4^{2018}}\)
Trừ theo vế:
$9D=4-\frac{2019}{4^{2018}}+\frac{2019}{4^{2019}}-\frac{1}{4^{2018}}$
$=4-\frac{6061}{4^{2019}}< 4$
$\Rightarrow D< \frac{4}{9}<\frac{4}{8}$ hay $D< \frac{1}{2}$ (đpcm)
2T
14 tháng 8 2019
\(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}\)
\(\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}\)
Đặt \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)
\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)
\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)
\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)
\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)
NM
2
KN
17 tháng 1 2020
Ta có: \(n^2>n^2-1=n^2-n+n-1=\left(n+1\right)\left(n-1\right)\)
Lúc đó:
\(B=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2019^3}\)
\(< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2018.2019.2020}\)
\(2B< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2018.2019.2020}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2018.2019}-\frac{1}{2019.2020}\)
\(=\frac{1}{2}-\frac{1}{2019.2020}< \frac{1}{2}\)
\(2B< \frac{1}{2}\Rightarrow B< \frac{1}{2^2}\)
Vậy \(B=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2019^3}< \frac{1}{2^2}\left(đpcm\right)\)
tương tự: https://lazi.vn/edu/exercise/tinh-tong-s-1-1-11-2-1-1-1-2-3-1-1-2-3-2018
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2019}\)
= \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2039190}\)
= \(2.\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4078380}\right)\)
= \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{2020}\right)\)
= \(2.\frac{1009}{2020}\)
= \(\frac{1009}{1010}\)