\(A=4+4^2+4^3+...+4^{100}\)

\(A=\left(4+\text{ }4^2\right)+\left(4^3+4^4\right)+...+\left(4^{99}+4^{100}\right)\)

\(A=\left(1+4\right).\left(4\right)+\left(1+4\right).\left(4^3\right)+...+\left(1+4\right).\left(4^{99}\right)\)

\(A=5.\left(4+4^3+4^5+...+4^{99}\right)\)

Vậy A chia hết cho 5

Các bạn nha!

    5+5^2 +5^3 +5^4...+5^99+5^100

= ( 5+5^2)+(5^3+5^4)+....+(5^99+5^100)

= 5(1+5)+5^3(1+5)+....+5^99(1+5)

=  5.6+5^3.6+....+5^99.6

= (5+5^3+....+5^99).6

Vì  (5+5^3+....+5^99).6 chia hết cho 6 nên 5+5^2 +5^3 +5^4...+5^99+5^100 chia hết cho 6.

1.

Ta có:

1/2 < 2/3

3/4 < 4/5

.............

99/100 < 100/101

=> 1/2*3/4*5/6*...*99/100 < 2/3*4/5*6/7*...*100/101

=> A < B

2.

\(A\cdot B=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)

\(A\cdot B=\frac{\left[1\cdot3\cdot5\cdot7\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot9\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)

3.

Vì A < B => A.A < A.B => A² < 1/101 < 1/100

Mà A² < 1/100 <=> A² < \(\frac{1}{10}^2\)=> A < 1/10

5^2+5^3+5^4+...+5^98+5^99=(5^2+5^3)+(5^4+5^5)+...+(5^98+5^99)=5^2.(1+5)+5^4.(1+5)+...+5^98.(1+5)=5^2.6+5^4.6+...+5^98.6=6.(5^2+5^4+...+5^98)=5^2+5^4+...+5^98 chia hết cho 6

A = 4 + 4² + 4³ + 4⁴ + ... + 4⁹⁹ + 4¹⁰⁰

A = ( 4 + 4² ) + ( 4³ + 4⁴ ) + ... + (4⁹⁹ + 4¹⁰⁰)

A = ( 4 + 4² ) + 4³(4 + 4² ) + .... + 4⁹⁹(4 + 4²)

A = 20 + 4³.20 + .... + 4⁹⁹.20

A = 20 ( 1 + 4³ + .... + 4⁹⁹ )

A = 4.5.(1 + 4^{3 + ...} + 4⁹⁹ ) ⋮ 5 ( đpcm )

\(A=4+4^2+4^3+4^4+...+4^{99}+4^{100}\)

\(A=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{99}+4^{100}\right)\)

\(A=4\left(4+1\right)+4^3\left(4+1\right)+...+4^{99}\left(4+1\right)\)

\(=5\left(4+4^3+...+4^{99}\right)\Rightarrow A⋮5\)

a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

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

Giúp mik vs ạ.Mik đag cần

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}< \frac{1}{2}\)

\(=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)