1,Tính nhanh
A=1/3+1/3^2+1/3^3+...+1/3^2007+1/3^2008
B=1/3+1/3^2+1/3^3+...+1/3^n-1+1/3^n ; n∈N*
2,Tính tổng
a,S=1/1.2.3+1/2.3.4+1/3.4.5+..+1/2006.2007.2008
b,S=1/1.2.3+1/2.3.4+1/3.4.5+..+1/n.(n+1).(n+2); n∈N*
K
Khách
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19 tháng 6 2022
a: \(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{n+1}}\)
\(\Leftrightarrow-\dfrac{2}{3}A=\dfrac{1}{3^{n+1}}-\dfrac{1}{3}\)
hay \(A=\left(\dfrac{1}{3^{n+1}}-\dfrac{1}{3}\right):\dfrac{-2}{3}=\dfrac{1-3^n}{3^{n+1}}\cdot\dfrac{3}{-2}=\dfrac{3^n-1}{3^n\cdot2}\)
b: \(\dfrac{1}{3}B=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2009}}\)
\(\Leftrightarrow B\cdot\dfrac{-2}{3}=\dfrac{1}{3^{2009}}-\dfrac{1}{3}=\dfrac{1-3^{2008}}{3^{2009}}\)
\(\Leftrightarrow B=\dfrac{3^{2008}-1}{3^{2009}}:\dfrac{2}{3}=\dfrac{3^{2008}-1}{2\cdot3^{2008}}\)
2 tháng 10 2020
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
\(\Rightarrow3A=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}\)
\(2A=3A-A\)
\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)
\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{2007}}-\frac{1}{3^{2008}}\)
\(=1-\frac{1}{3^{2008}}\)
\(2A=1-\frac{1}{3^{2008}}\Rightarrow A=\frac{1-\frac{1}{3^{2008}}}{2}\)
F
2 tháng 10 2020
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\)
\(\Leftrightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)
\(\Leftrightarrow2A=1-\frac{1}{3^{2008}}\)
\(\Leftrightarrow2A=\frac{3^{2008}-1}{3^{2008}}\)
\(\Leftrightarrow A=\frac{3^{2008}-1}{3^{2008}}\div2\)
\(\Leftrightarrow A=\frac{3^{2008}-1}{2.3^{2008}}\)
NL
2
LD
17 tháng 7 2017
\(\frac{M}{N}=\frac{\frac{1}{2007}+\frac{2}{2006}+......+\frac{2006}{2}+\frac{2007}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2006}+\frac{1}{2007}}\)
\(\frac{M}{N}=\frac{\frac{1}{2007}+1+\frac{2}{2006}+1+.......+\frac{2007}{1}+1+\frac{2008}{2008}-2008}{\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+.....+\frac{1}{2}}\)
\(\frac{M}{N}=\frac{\frac{2008}{2007}+\frac{2008}{2006}+....+\frac{2008}{1}+\frac{2008}{2008}-2008}{\frac{1}{2008}+........+\frac{1}{2}}\)
đến đây là ra rùi ha
AD
1
16 tháng 10 2015
Đặt A=1+a+a^2+a^3+...+a^n
a*A=a+a^2+a^3+a^4+...+an+1
a*A+1=1+a+a^2+a^3+...+a^n+an+1=A+an+1
a*A-A=an+1-1
(a-1)A=an+1-1
A=(an+1-1)/(a-1)
28 tháng 12 2015
\(M:N=\frac{\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
Ta có tử số bằng: 2008+2007/2+2006/3+2005/4+…..+2/2007+1/2008
(Phân tích 2008 thành 2008 con số 1 rồi đưa vào các nhóm)
= (1 + 2007/2) + (1 + 2006/3) + (1 + 2005/4) +... + (1 + 2/2007) + ( 1 + 1/2008) + (1)
= 2009/2 + 2009/3 + 2009//4 + ……. + 2009/2007 + 2009/2008 + 2009/2009
= 2009 x (1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009)
Mẫu số: 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009
\(\Rightarrow M:N=\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)
GT
1.chứng minh rằng : 1^3+2^3+3^3+...+n^3 chia hết 1+2+3+...+n
2.tìm x , 1/3+1/6+...+2/x(x+1)=2005/2007
1
5 tháng 7 2015
Cái bài 2 nhân với 1 là 2/2 nên nhân cả tử cả mẫu với 2 ra 6=2*3
12=3*4
.........
Còn lại tự tính
Nếu ra kết quả đúng thì cho **** nhé
8 tháng 1 2016
A=2008+2007/2+2006/3+2005/4+...+2/2007+1/2008
1/2+1/3+1/4+1/5+...+1/2007+1/2008
=(1+2007/2)+(1+2006/3)+(1+2005/4)+...+(1+2/2007)+(1+1/2008)
1/2+1/3+1/4+...+1/2008
=2009(1/2+1/3+1/4+...+1/2008)
1/2+1/3+1/4+..+1/2008
=2009
A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
3A-A= \(1-\frac{1}{3^{2008}}\)
B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}+\frac{1}{3^n}\)
3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-2}}+\frac{1}{3^{n-1}}\)
3B - B = \(1-\frac{1}{3^n}\)