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\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\left(1\right)\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\left(2\right)\)
Lấy (2) - (1) ta được:\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\right)\)
\(\Leftrightarrow2A=1-\frac{1}{3^{100}}\)
\(\Leftrightarrow A=\left(\frac{3^{100}-1}{3^{100}}\right):2\)
\(\Leftrightarrow A=\frac{3^{100}-1}{2.3^{100}}\)
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Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
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3A = 1 + 1/3 + 1/3^2 + ... + 1/3^199
3A - A = ( 1 + 1/3 + 1/3^2 + ... + 1/3^99 ) - ( 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^100 )
2A = 1 - 1/3^100
A = ( 1 - 1/3^100 ) / 2
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(A=\frac{3^{100}-1}{3^{100}.2}\)
mk chỉ làm được đến đây thôi
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A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100
Ta đổi A = 2-1+1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100
A= 2 - 1 - 1/100 =200/100 -100/100 - 1/100
A= 99/100
Cảm ơn bạn Kudo Shinichi, nhưng
1=2-1 ->ok
1/2=1-1/2 ->ok
1/3=1/2-1/3 -> sai
vì 1/2-1/3=1/6
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ĐẶT A=\(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\frac{1}{3}A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)
\(\frac{1}{3}A-A=\frac{1}{3^{2006}}-\frac{1}{3^0}\)
\(\frac{-2}{3}A=\frac{1}{3^{2006}}-\frac{1}{3^0}\)
\(A=\frac{\frac{1}{3^{2006}}-1}{\frac{-2}{3}}\)
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*) A=1+6+11+16+21+....+101
Dãy trên có: \(\left(101-1\right):5+1=21\)(số số hạng)
\(\Rightarrow A=\frac{\left(101+1\right)\cdot21}{2}=1071\)
*) Đặt C=\(1^2+2^2+3^2+....+98^2=1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\)
\(\Rightarrow B-C=\left(1\cdot2+2\cdot3+3\cdot4+....+98\cdot99\right)-\left(1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\right)\)
\(=\left(1\cdot2-1\cdot1\right)+\left(2\cdot3-2\cdot2\right)+\left(3\cdot4-3\cdot3\right)+.....+\left(98\cdot99-98\cdot98\right)\)
\(=1\left(2-1\right)+2\left(3-2\right)+3\left(4-3\right)+....+98\left(99-1\right)\)
\(=1\cdot1+2\cdot1+3\cdot1+....+98\cdot1\)
\(=1+2+3+....+98\)
\(=\frac{\left(98+1\right)\cdot98}{2}=4851\)
A = 1 + 6 + 11 + 16 +21 +... + 101
Số chữ số của tổng A là :
( 101 - 1 ) : 5 + 1 = 21 (số)
Tổng A = 1 + 6 + ... + 101 = (101 + 1) . 21 : 2 = 1071
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\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+....+\frac{10}{1400}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+.....+\frac{5}{25.28}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\frac{3}{14}=\frac{5}{14}\)
a)Đặt A=Tổng trên, ta có:
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(2A=2+1+...+\frac{1}{2^{99}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(A=2-\frac{1}{2^{100}}\)
b)có đứa làm rồi
c)Đặt C=Tổng trên
\(3C=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{300}}\right)\)
\(3C=1+\frac{1}{3}+...+\frac{1}{3^{299}}\)
\(3C-C=\left(1+\frac{1}{3}+...+\frac{1}{3^{299}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{300}}\right)\)
\(2C=1-\frac{1}{3^{300}}\)
\(C=\frac{1-\frac{1}{3^{300}}}{2}\)
\(s=\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}\)
3S = \(1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\)
=> S = \(\frac{3S-S}{2}=\frac{1-\frac{1}{3^{100}}}{2}\)