C=1/2+1/3+...+1/99+1/100
D=99/1+98/2+...+1/99
tìm C/D
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Ta thấy:
\(A=1\cdot3+2\cdot4+...+97\cdot99+98\cdot100\)
\(A=1\cdot\left(1+2\right)+2\cdot\left(1+3\right)+...+97\cdot\left(1+98\right)+98\cdot\left(1+99\right)\)
\(A=\left(1+1\cdot2\right)+\left(2+2\cdot3\right)+...+\left(97+97\cdot98\right)+\left(98+98\cdot99\right)\)
\(A=\left(1+2+...+97+98\right)+\left(1\cdot2+2\cdot3+...+97\cdot98+98\cdot99\right)\)
Đặt \(B=1+2+...+97+98\) ; \(C=1\cdot2+2\cdot3+...+97\cdot98+98\cdot99\). Khi đó: \(A=B+C\)
* Do số các số hạng của tổng B là: ( 98 - 1 ) : 1 + 1 = 98 ( số hạng ) nên:
\(B=1+2+...+97+98=\frac{\left(98+1\right)\cdot98}{2}=99\cdot49=4851\)
* Ta thấy:
\(C=1\cdot2+2\cdot3+...+97\cdot98+98\cdot99\)
\(\Rightarrow3\cdot C=1\cdot2\cdot3+2\cdot3\cdot3+...+97\cdot98\cdot3+98\cdot99\cdot3\)
\(\Rightarrow3\cdot C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+97\cdot98\cdot\left(99-96\right)+98\cdot99\cdot\left(100-97\right)\)
\(\Rightarrow3\cdot C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+97\cdot98\cdot99-96\cdot97\cdot98+98\cdot99\cdot100-97\cdot98\cdot99\)
\(\Rightarrow3\cdot C=98\cdot99\cdot100\)
\(\Rightarrow C=\frac{98\cdot99\cdot100}{3}\)
\(\Rightarrow C=98\cdot33\cdot100\)
\(\Rightarrow C=323400\)
Vậy: \(A=B+C=4851+323400=328251\)
a) \(227+50+23=\left(227+23\right)+50=250+50=300\)
b) \(135+360+65+40=\left(135+65\right)+\left(360+40\right)=200+400=600\)
c) \(1+2+3+4+5+...+97+98+99+100\)
\(=\left(100+1\right)+\left(99+2\right)+...+\left(50+51\right)\)
\(=101+101+101+...+101\)
\(=101\cdot50\)
\(\Leftrightarrow5050\)
d) \(115\cdot13-13\cdot15=13\cdot\left(115-15\right)=13\cdot100=1300\)
e) \(50-49+48-47+...+4-3+2-1\)
\(=\left(50-49\right)+\left(48-47\right)+...+\left(2-1\right)\)
\(=1+1+1+1+..+1\)
\(=1\cdot25\)
\(=25\)
f) \(30\cdot40\cdot50\cdot60=10\cdot3+10\cdot4+10\cdot5+10\cdot6\)
\(=10\cdot10\cdot10\cdot10\cdot3\cdot4\cdot5\cdot6\)
\(=10000\cdot360\)
\(=3600000\)
g) \(27\cdot36+27\cdot64=27\cdot\left(36+64\right)=27\cdot100=2700\)
h) \(5\cdot2^2-18:3=5\cdot4-18:3=20-6=14\)
i) \(13\cdot17-256:16+14:7-2021^0\)
\(=13\cdot17-4^4:4^2+2-1\)
\(=13\cdot17-16+2-1\)
\(=13\cdot17-17\)
\(=17\cdot\left(13-1\right)\)
\(=204\)
j) \(7^2-36:3=49-12=37\)
\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+\frac{1}{98\cdot97}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(\Rightarrow C=\frac{1}{100}-1+\frac{1}{100}\)
\(\Rightarrow C=\left(\frac{1}{100}+\frac{1}{100}\right)-1\)
\(\Rightarrow C=\frac{1}{50}-1\)
\(\Rightarrow C=\frac{-49}{50}\)
=1/100-(1/1x2+1/2x3+...+1/99x100)
=1/100-(1-1/2+1/2-1/3+...+1/99-1/100)
=1/100-(1-1/100)
=1/100-1+1/100
=2/100-1
=-49/50
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
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