= 3/100

a) Ta có:

(n-1)/n < n/(n+1)

vì (n-1).(n+1)=n²-1 < n²

=>

1/2 < 2/3

3/4 < 4/5

....

99/100 < 100/101

Vậy A < B

b). Ta lại có:

A.B = 1/2 . 2/3 . 3/4 . 4/5 .... . 99/100 . 100/101 = 1/100

Mà A<B => A.A<A.B=1/100

=> A² < 1/100

=> A < 1/10<1

A < 1/10<1

A < 1/10<1

Đặt \(A=1+2+2^2+2^3+...+2^{100}\\ 2A=2+2^2+2^3+2^4+...+2^{101}\\ 2A-A=2+2^2+2^3+2^4+...+2^{101}-1-2-2^2-2^3-...-2^{100}\\ A=2^{101}-1\)

\(\Rightarrow2^{100}-\left(2^{100}-1\right)=2^{100}-2^{101}+1\)

#)Giải :

Đặt \(A=1.2+2.3+3.4+...+99.100\)

\(3A=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(3A=0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51\)

\(3A=49.50.51=124950\)

\(\Leftrightarrow A=\frac{124950}{3}=41650\)

Mình sửa lại đề vì sai :  1 x 2 + 2 x 3 + ... + 99 x 100

Đặt A = 1 x 2 + 2 x 3 + ... + 99 x 100

  => 3 x A = 3 x (1 x 2 + 2 x 3 + ... + 99 x 100)

  => 3 x A = 1 x 2 x 3 + 2 x 3 x 3 + ... + 99 x 100 x 3

  => 3 x A = 1 x 2 x 3 + 2 x 3 x (4 - 1) + ... + 99 x 100 x (101 - 98)

  => 3 x A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ... + 99 x 100 x 101 - 98 x 99 x 100

  => 3 x A = 99 x 100 x 101

  => 3 x A = 999 900

  => A       = 999 900 : 3

  => A       = 333 300

Vậy 1 x 2 + 2 x 3 + ... + 99 x 100 = 333 300

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

3A = 99.100.101

A = 99.100.101 : 3 = 333300

????

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