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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2004.2005.2006}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2004.2005}-\frac{1}{2005.2006}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2005.2006}\right)\)
\(=\frac{1}{4}-\frac{1}{2.2005.2006}\)
Sửa lại cái dòng này một tí:
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.200}-\frac{1}{200.201}\)
Còn lại đúng hết! Không cần lo
a) Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{199.200.201}\). Ta xét:
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{1}{1.2.3}\); \(\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{2.3.4}\); \(\frac{1}{3.4}-\frac{1}{4.5}=\frac{1}{3.4.5}\);.......;\(\frac{1}{99.200}-\frac{1}{200.201}=\frac{1}{99.100.101}\)
Qua công thức trên ta rút ra tổng quát ( nói thêm cho dễ hiểu)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{199.200.201}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{199.200.201}\)
Ta thấy: \(-\frac{1}{2.3}+\frac{1}{2.3}=0\);\(-\frac{1}{3.4}+\frac{1}{3.4}=0\); . . . . .
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{200.201}=\frac{1}{2}-\frac{1}{40200}\)
\(\Rightarrow A=\frac{\frac{1}{2}-\frac{1}{40200}}{2}\)
1)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)=\frac{1}{3}+\left(-1\right)+1=\frac{1}{3}\)
Sửa đề chút nha
\(\frac{x}{2}=\frac{1}{1.2.3}+....+\frac{1}{98.99.100}\)
Ta có công thức tổng quát \(\frac{1}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{2}\left(\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\right)\)
\(\Rightarrow\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
Áp dụng vào tổng ta có
\(\frac{x}{2}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{98.99}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{99.100}=\frac{4949}{9900}\)
\(\Rightarrow x=\frac{4949}{4950}\)
Đặt biểu thức là A, ta có:
3A= \(\frac{3}{1\times2\times3}+\frac{3}{2\times3\times4}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
3A = \(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
3A = \(\frac{1}{1\times2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=>A = \(\frac{3}{2}\) - \(\frac{3}{\left(n+1\right)\left(n+2\right)}\)
Phần b bạn làm tương tự
S=1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +...+ 1/2010.2011 - 1/2011.2012
S=1/1.2 - 1/2011.2012<1/2
=>S<P
s=1/1*2-1/2*3+1/2*3-1/3*4+....+1/2009*2010-1/210*2011
=1/1*2-1/2010*2011
<1/1*2
\(S=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2009\cdot2010\cdot2011}\)
\(S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{2009\cdot2010}-\frac{1}{2010\cdot2011}\)
\(S=\frac{1}{1\cdot2}-\frac{1}{2010\cdot2011}\)
\(S=\frac{1}{2}-\frac{1}{2010\cdot2011}< \frac{1}{2}\)
=> S < P
a ) Co :
1/1.2 - 1/2.3 = 2/1.2.3
1/2.3 - 1/3.4 = 2/2.3.4
...
1/37.38 - 1/38.39 = 2/37.38.39
=> 2M = 2/1.2.3 + 2/2.3.4 + ... + 2/37.38.39
=> 2M = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/37.38 - 1/38.39
=> 2M = 1/2 - 1/1482
=> 2M = 370/741
=> M = 185/741
B ) A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8
3A = 1 + 1/3 + 1/3^2 + ... + 1/3^7
3A - A = ( 1 + 1/3 + 1/3^2 + ... + 1/3^7 ) - ( 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8 )
2A = 1 - 1/3^8
A = ( 1 - 1/3^8 ) / 2
Có \(\frac{1}{1.2.3}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)
\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)
...
\(\frac{1}{17.18.19}=\frac{1}{2}\left(\frac{1}{17.18}-\frac{1}{18.19}\right)\)
=>\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{17.18.19}\)=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{17.18}-\frac{1}{18.19}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{18.19}\right)=\frac{1}{2}.\frac{1}{2}-\frac{1}{2}.\frac{1}{18.19}< \frac{1}{4}\)
Đặt A=(đã cho).
=>2A=2/1*2*3+2/2*3*4+2/3*4*5+...+2/37*38*39.
=>2A=1/1*2-1/2*3+1/2*3-1/3*4+...+1/37*38-1/38*39.
=>2A=1/1*2=1/38*39.
Đến đây tự bấm máy nha.
tk mk nha.
chắc chắn đúng,nay mk làm bài này.
-chúc ai tk mk học giỏi-
\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2004.2005.2006}\)
\(=\frac{2}{1.2}-\frac{2}{2.3}+\frac{2}{2.3}-\frac{2}{3.4}+...+\frac{2}{2004.2005}-\frac{2}{2005.2006}\)
\(=\frac{2}{1.2}-\frac{2}{2005.2006}\)
\(=1-\frac{1}{2011015}\)
\(=\frac{2011015}{2011015}-\frac{1}{2011015}\)
\(=\frac{2011014}{2011015}\)
Cbht