a) P = 1 + 3 + 3² + ... + 3¹⁰¹

= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)

= 13 + 3³.(1 + 3 + 3²) + ... + 3⁹⁹.(1 + 3 + 3²)

= 13 + 3³.13 + ... + 3⁹⁹.13

= 13.(1 + 3³ + ... + 3⁹⁹) ⋮ 13

Vậy P ⋮ 13

b) B = 1 + 2² + 2⁴ + ... + 2²⁰²⁰

= (1 + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰)

= 21 + 2⁶.(1 + 2² + 2⁴) + ... + 2²⁰¹⁶.(1 + 2² + 2⁴)

= 21 + 2⁶.21 + ... + 2²⁰¹⁶.21

= 21.(1 + 2⁶ + ... + 2²⁰¹⁶) ⋮ 21

Vậy B ⋮ 21

c) A = 2 + 2² + 2³ + ... + 2²⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁷ + 2¹⁸ + 2¹⁹ + 2²⁰)

= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2¹⁶.(2 + 2² + 2³ + 2⁴)

= 30 + 2⁴.30 + ... + 2¹⁶.30

= 30.(1 + 2⁴ + ... + 2¹⁶)

= 5.6.(1 + 2⁴ + ... + 2¹⁶) ⋮ 5

Vậy A ⋮ 5

d) A = 1 + 4 + 4² + ... + 4⁹⁸

= (1 + 4 + 4²) + (4³ + 4⁴ + 4⁵) + ... + (4⁹⁷ + 4⁹⁸ + 4⁹⁹)

= 21 + 4³.(1 + 4 + 4²) + ... + 4⁹⁷.(1 + 4 + 4²)

= 21 + 4³.21 + ... + 4⁹⁷.21

= 21.(1 + 4³ + ... + 4⁹⁷) ⋮ 21

Vậy A ⋮ 21

e) A = 11⁹ + 11⁸ + 11⁷ + ... + 11 + 1

= (11⁹ + 11⁸ + 11⁷ + 11⁶ + 11⁵) + (11⁴ + 11³ + 11² + 11 + 1)

= 11⁵.(11⁴ + 11³ + 11² + 11 + 1) + 16105

= 11⁵.16105 + 16105

= 16105.(11⁵ + 1)

= 5.3221.(11⁵ + 1) ⋮ 5

Vậy A ⋮ 5

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

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

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

S=(1+2)+...+2^6(1+2)=3(1+...+2^6)⋮3

a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)

\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)

Vậy ta có biểu thức:

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)

Vậy B < 1 (đpcm)

Giải:

a) Ta có:

1/2²=1/2.2 < 1/1.2

1/3²=1/3.3 < 1/2.3

1/4²=1/4.4 < 1/3.4

1/5²=1/5.5 < 1/4.5

1/6²=1/6.6 < 1/5.6

1/7²=1/7.7 < 1/6.7

1/8²=1/8.8 <1/7.8

⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8

   B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8

   B<1/1-1/8

   B<7/8

mà 7/8<1

⇒B<7/8<1

⇒B<1

b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46

   S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

   S=1/1-1/46

   S=45/46

Vì 45/46<1 nên S<1

Vậy S<1

Chúc bạn học tốt!