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Đặt tổng của 2005 số hạng đầu tiên của dãy là S
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{2005.2006}\)
\(S=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+..+\frac{2006-2005}{2005.2006}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{2005}-\frac{1}{2006}\)
\(S=1-\frac{1}{2006}=\frac{2005}{2006}\)
A = 1.2 + 2.3 + 3.4 + ... + 2005.2006
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2005.2006.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2005.2006.(2007 - 2004)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2005.2006.2007 - 2004.2005.2006
3A = 2005.2006.2007
A = 2690738070
Có A=1.2+2.3+3.4+...+2005.2006
3A=1.2.3+2.3.3+3.4.3+....+2005.2006.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+....+2005.2006.(2007-2004)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+2005.2006.2007-2004.2005.2006
3A=2005.2006.2007
A=(2005.2006.2007):3
Vậy A=....
Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.
=1/1-1/2022
=2021/2022
Dễ thôi!
Ta có: 1/1.2 = 1/1 - 1/2 ; 1/2.3 = 1/2 - 1/3 ; 1/3.4 = 1/3 - 1/4 ; ...;1/99.100 = 1/99 - 1/100
Như vậy thì bài toán trên = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/99 - 1/100
Vậy tổng trên là:
1 - 1/100
= 99/100
tk nha
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{1000}{1000}-\frac{1}{1000}+\frac{1000}{1000}\)
\(=\frac{1999}{1000}\)
Tham khảo nhé~
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{149.150}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{149}-\dfrac{1}{150}\)
\(A=\dfrac{1}{1}-\dfrac{1}{150}\)
\(A=\dfrac{150}{150}-\dfrac{1}{150}\)
\(A=\dfrac{149}{150}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{49.50}\)
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
A=1-\(\frac{1}{50}\)
A=\(\frac{49}{50}\)
A=1/1.2+1/2.3+1/3.4+..+1/99.100
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100
=99/100
1/1.2+1/2.3+1/3.4+...+1/2005.2006=(1-1/2)+(1/2-1/3)+...+(1/2005-1/2006)=1-1/2+1/2-1/3+...+1/2005-1/2006
=1-(1/2-1/2)+...-1/(1/2005-1/2005)-1/2006=1-1/2006=2005/2006
k mình nha
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