Thu gọn: \(\dfrac{1}{3}\)+ \(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}^{ }\)+...+ \(\dfrac{100}{3^{100}}\)
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DT
22 tháng 8 2017
a) Đặt \(C=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)
\(\Rightarrow5C=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{99}}\)
\(\Rightarrow5C-C=1-\dfrac{1}{5^{100}}\Rightarrow4C=1-\dfrac{1}{5^{100}}\Rightarrow C=\dfrac{1-\dfrac{1}{5^{100}}}{4}\)
\(\Rightarrow A=8.5^{100}.\dfrac{1-\dfrac{1}{5^{100}}}{4}+1=2.\left(5^{100}-1\right)+1=2.5^{100}-2+1=2.5^{100}-1\)
DT
22 tháng 8 2017
b)\(B=\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)
\(B=4.\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)\)
Đặt \(\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)=D\)
\(\Rightarrow3D=1-\dfrac{1}{3}+...-\dfrac{1}{3^{99}}\)
\(\Rightarrow3D+D=1-\dfrac{1}{3^{100}}\)
\(\Rightarrow D=\dfrac{1-\dfrac{1}{3^{100}}}{4}\)
VN
1
15 tháng 12 2017
\(A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}\)
\(3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{100}{3^{99}}\)
\(3A-A=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}\right)\)
\(2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
Đặt: \(L=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(3L=3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\)
\(3L-L=\left(3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}...+\dfrac{1}{3^{99}}\right)\)
\(2L=3-\dfrac{1}{3^{99}}\)
\(L=\dfrac{3}{2}-\dfrac{1}{3^{99}.2}\)
Hay: \(2A=\dfrac{3}{2}-\dfrac{1}{3^{99}.2}-\dfrac{100}{3^{100}}\)
\(A=\dfrac{3}{4}-\dfrac{1}{3^{99}.4}-\dfrac{100}{3^{100}.2}\)
16 tháng 3 2023
q=1/3; u1=2/3
\(S_{100}=\dfrac{\dfrac{2}{3}\cdot\left(\dfrac{1}{3^{100}}-1\right)}{\dfrac{1}{3}-1}=-\dfrac{1}{3^{100}}+1=\dfrac{-1+3^{100}}{3^{100}}\)
11 tháng 6 2021
Với n\(\in N\)* có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow\)\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) (*)
a) Áp dụng (*) vào T
\(\Rightarrow T=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
b) Có \(VT=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)\(=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)
\(\Leftrightarrow\sqrt{n+1}=5\Leftrightarrow n=24\) (tm)
Vậy n=24.
14 tháng 11 2023
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 4 2022
Xét hàm:
\(f\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+...+\dfrac{1}{x^{100}}\)
\(\Rightarrow f'\left(x\right)=-\dfrac{1}{x^2}-\dfrac{2}{x^3}-\dfrac{3}{x^4}-...-\dfrac{100}{x^{101}}=-P\) (1)
Mặt khác \(f\left(x\right)\) là tổng cấp số nhân với \(\left\{{}\begin{matrix}n=100\\u_1=\dfrac{1}{x}\\q=\dfrac{1}{x}\end{matrix}\right.\)
\(\Rightarrow f\left(x\right)=u_1.\dfrac{1-q^{100}}{1-q}=\dfrac{1}{x}.\dfrac{1-\dfrac{1}{x^{100}}}{1-\dfrac{1}{x}}=\dfrac{1-\dfrac{1}{x^{100}}}{x-1}=\dfrac{x^{100}-1}{x^{101}-x^{100}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{\left(x^{100}-1\right)'\left(x^{101}-x^{100}\right)-\left(x^{101}-x^{100}\right)'\left(x^{100}-1\right)}{\left(x^{101}-x^{100}\right)^2}=-\dfrac{x^{101}-101x^{100}+100}{x^{101}\left(x-1\right)^2}\) (2)
(1);(2) \(\Rightarrow P=\dfrac{x^{101}-101x^{100}+100}{x^{101}\left(x-1\right)^2}\)
T
8 tháng 7 2019
Em thử thôi chứ ko chắc đâu:((
Xét dạng tổng quát \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}\)
Suy ra \(A=\frac{1}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{100-\sqrt{100}}{100}\)
AT
11 tháng 6 2021
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)
11 tháng 6 2021
cả 2 ý bạn trục căn thức ở mấu là xong nhé:
vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy
Đặt \(A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)
\(\Rightarrow3A-A=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{100}}\right)\)
\(\Rightarrow2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\dfrac{100}{3^{100}}\)
Đặt \(B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3B=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3B-B=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2B=3-\dfrac{1}{3^{99}}\Rightarrow B=\dfrac{3}{2}-\dfrac{1}{3^{99}.2}\)
Do đó: \(2A=\dfrac{3}{2}-\dfrac{1}{3^{99}.2}-\dfrac{100}{3^{100}}=\dfrac{3^{101}}{3^{100}.2}-\dfrac{3}{3^{100}.2}-\dfrac{200}{3^{100}.2}=\dfrac{3^{101}-203}{3^{100}.2}\Rightarrow A=\dfrac{3^{101}-203}{3^{100}.4}\)
Vậy...