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8 tháng 1 2016

S = 12 + 22 + 32 + ..... + 1002

= 1.(2-1) + 2.(3-1) + .... + 100.(101-1)

= 1.2 + 2.3 + ....... + 100.101 - (1+2+3+...+100)

Đặt A = 1.2 + 2.3 + ... + 100.101

3A = 1.2.(3-0) + 2.3.(4-1) + .... + 100.101.(102-99)

3A = 1.2.3  + 2.3.4-1.2.3 + ... +100.101.102-99.100.101

3A = 100.101.102

A = 100.101.102 : 3 = 33367034

S = 33367034 - 5050

S = 33361984

 

8 tháng 1 2016

S = 12 + 22 + 32 + ..... + 1002

= 1.(2-1) + 2.(3-1) + .... + 100.(101-1)

= 1.2 + 2.3 + ....... + 100.101 - (1+2+3+...+100)

Đặt A = 1.2 + 2.3 + ... + 100.101

3A = 1.2.(3-0) + 2.3.(4-1) + .... + 100.101.(102-99)

3A = 1.2.3  + 2.3.4-1.2.3 + ... +100.101.102-99.100.101

3A = 100.101.102

A = 100.101.102 : 3 = 33367034

S = 33367034 - 5050

S = 33361984

 

10 tháng 9 2015

C1: \(A=\left(\frac{36}{6}-\frac{4}{6}+\frac{3}{6}\right)-\left(\frac{150}{30}+\frac{50}{30}-\frac{45}{30}\right)-\left(\frac{18}{6}-\frac{14}{6}+\frac{15}{6}\right)\)

\(=\frac{35}{6}-\frac{155}{30}-\frac{19}{6}=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=-\frac{15}{6}=-2\frac{1}{2}\)

C2: \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

\(=\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)

\(=-2-0-\frac{1}{2}=-2\frac{1}{2}\)

8 tháng 2 2018

\(B=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+.......+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)

\(\Leftrightarrow2B=1+\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+........+\dfrac{98}{2^{99}}+\dfrac{99}{2^{100}}\)

\(\Leftrightarrow2B-B=\left(1+\dfrac{1}{2^2}+\dfrac{2}{2^3}+........+\dfrac{99}{2^{100}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+......+\dfrac{100}{2^{100}}\right)\)

\(\Leftrightarrow B=\dfrac{1}{2}+\dfrac{1}{2^2}+..........+\dfrac{1}{2^{100}}-\dfrac{100}{2^{100}}\)

Đặt :

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{99}}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+......+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{100}}\)

\(\Leftrightarrow B=1-\dfrac{1}{2^{100}}-\dfrac{100}{2^{100}}\)

\(\Leftrightarrow B=\dfrac{2^{100}-101}{2^{100}}\)

9 tháng 2 2018

thank you

12 tháng 4 2017

A=1+3/2^3+4/2^4+5/2^5+...100/2^100
1/2*A = 1/2 + 3/2^4 + 4/2^5 +....+ 99/2^100 + 100/2^101

A- A/2 = 1/2A =1/2 + 3/2^3 + 1/2^4 +...+1/2^100 - 100/2^101=

= [1/2+1/2^2 +1/2^3 +...+1/2^100] -100/2^101 (Do 3/2^3 = 1/2^2 +1/2^3)

=[1-(1/2)^101]/(1-1/2) -100/2^101 =

=(2^101 -1)/2^100 - 100/2^101

=> A= (2^101 -1)/2^99 - 100/2^100