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16 tháng 7 2017

a) A = 2 + 22 + 23 + ... + 2100

2A = 22 + 23 + 24 + ... + 2101

2A - A = (22 + 23 + 24 + ... + 2101) - (2 + 22 + 23 + ... + 2100)

A = 2101 - 2

b) B = 1 + 3 + 32 + ... + 3255

3B = 3 + 32 + 33 + ... + 3256

3B - B = (3 + 32 + 33 + ... + 3256) - (1 + 3 + 32 + ... + 3255)

2B = 3256 - 1

B = \(\frac{3^{256}-1}{2}\)

c) C = 1 + 4 + 42 + ... + 4100

4C = 4 + 42 + 43 + ... + 4101

4C - C = (4 + 42 + 43 + ... + 4101) - (1 + 4 + 42 + ... + 4100)

3C = 4101 - 1

C = \(\frac{4^{101}-1}{3}\)

d) D = 1 + 5 + 52 + ... + 51000

5D = 5 + 52 + 53 + ... + 51001

5D - D = (5 + 52 + 53 + ... + 51001) - (1 + 5 + 52 + ... + 51000)

4D = 51001 - 1

D = \(\frac{5^{1001}-1}{4}\)

7 tháng 11 2018

cứ tổng hai số hạng sẽ chia hết cho 3 nhé 

7 tháng 11 2018

\(A=1+2+2^2+2^3+...+2^{11}\)

\(A=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{10}+2^{11}\right)\)

\(A=3+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)\)

\(A=3+2^2.3+...+2^{10}.3\)

\(A=3\left(1+2^2+...+2^{10}\right)\)

\(\Rightarrow A⋮3\)                  

Vậy \(A⋮3\)

  !!!

30 tháng 4 2019

\(\frac{B}{A}=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\frac{B}{A}=\frac{1+\left[\frac{1}{99}+1\right]+\left[\frac{2}{98}+1\right]+\left[\frac{3}{97}+1\right]+...+\left[\frac{98}{2}+1\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\frac{B}{A}=\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\frac{B}{A}=\frac{100\cdot\left[\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right]}{\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right]}=100\)

Vậy : \(\frac{B}{A}=100\)

30 tháng 4 2019

Ta có:

\(B=\frac{1}{99}+\frac{2}{98}+...+\frac{99}{1}\)

\(=\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)+...+\left(1+\frac{98}{2}\right)+1\)

\(=\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}\)

\(=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)

\(=100.A\)

\(\Rightarrow\frac{B}{A}=100\)

9 tháng 9 2016

a) đặt tên biểu thức là A. Ta có :

A =  1.2+2.3+3.4+...+99.100

3A = 1.2.3+2.3.3+3.4.3+...+99.100.3

3A = 1.2.3 + 2.3.(4-1 ) + 3.4.(5-2) + ... + 99.100.(101-98)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

A = 99.100.101 : 3

A = 333300

b) đặt tên biểu thức là B ta có :

B= 1.2+2.3+3.4+...+n.(n+1)

3B = 1.2.3+2.3.3+3.4.3+...+n.(n+1).3

3B = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ... + n.(n+1).[ (n+2) - ( n -1 ) ]

3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + n.(n+1).(n+2) - (n-1).n.(n+1)

B = n.(n+1).(n+2) : 3

9 tháng 9 2016

\(A=1\cdot2+2\cdot3+...+99\cdot100\)

\(3\cdot A=1\cdot2\cdot3+2\cdot3\cdot3+...+99\cdot100\cdot3\)

\(3\cdot A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(3\cdot A=1\cdot2\cdot3+2\cdot3\cdot4+...+99\cdot100\cdot101-1\cdot2\cdot3-...-98\cdot99\cdot100=\)

\(3\cdot A=99\cdot100\cdot101\)

\(A=99\cdot100\cdot101\div3=333300\)

CCâu b tương tự

14 tháng 3 2017

Ta có:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
\(\Rightarrow3\cdot A=3\cdot\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\right)\)
\(\Rightarrow3\cdot A=3\cdot\frac{1}{3}+3\cdot\frac{2}{3^2}+3\cdot\frac{3}{3^3}+...+3\cdot\frac{100}{3^{100}}+3\cdot\frac{101}{3^{101}}\)
\(\Rightarrow3\cdot A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\)
\(\Rightarrow3\cdot A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\right)\)
\(\Rightarrow2\cdot A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}-\frac{1}{3}-\frac{2}{3^2}-\frac{3}{3^3}-...-\frac{100}{3^{100}}-\frac{101}{3^{101}}\)
\(\Rightarrow2\cdot A=1+\left(\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+...+\left(\frac{101}{3^{100}}-\frac{100}{3^{100}}\right)-\frac{101}{3^{101}}\)
\(\Rightarrow2\cdot A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
Khi đặt \(S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\) thì ta sẽ có 2 điều:
- Điều 1: Khi đó:
\(2\cdot A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
\(\Rightarrow2\cdot A=S-\frac{101}{3^{101}}\)
\(\Rightarrow2\cdot A< S\)    ( 1 )
Điều 2: Khi đó:
\(S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3\cdot S=3\cdot\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow3\cdot S=3\cdot1+3\cdot\frac{1}{3}+3\cdot\frac{1}{3^2}+...+3\cdot\frac{1}{3^{100}}\)
\(\Rightarrow3\cdot S=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3\cdot S-S=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow2\cdot S=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-1-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)
\(\Rightarrow2\cdot S=3+\left(1-1\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{99}}-\frac{1}{3^{99}}\right)-\frac{1}{3^{100}}\)
\(\Rightarrow2\cdot S=3+0+0+0+...+0-\frac{1}{3^{100}}\)
\(\Rightarrow2\cdot S=3-\frac{1}{3^{100}}\)
Do \(3-\frac{1}{3^{100}}< 3\) nên:
\(\Rightarrow2\cdot S< 3\)
\(\Rightarrow S< \frac{3}{2}\)    ( 2 )
Từ ( 1 ) và ( 2 ), theo tính chất bắc cầu suy ra:
\(2\cdot A< \frac{3}{2}\)
\(\Rightarrow A< \frac{3}{2}:2\)
\(\Rightarrow A< \frac{3}{2\cdot2}\)
\(\Rightarrow A< \frac{3}{4}\)    ( đpcm )