a) Ta có:

$S=1-3+3^2-3^3+\cdots+3^{98}-3^{99}$

$=(1-3)+(3^2-3^3)+\cdots+(3^{98}-3^{99})$

$=-2-2\cdot3^2-\cdots-2\cdot3^{98}$

$=-2(1+3^2+3^4+\cdots+3^{98})$

$=-2(1+9+9^2+\cdots+9^{49})$

$=-2\cdot\dfrac{9^{50}-1}{9-1}$

$=-\dfrac{9^{50}-1}{4}$

Muốn $S$ chia hết cho $-20$ thì cần:

$S\vdots20$

Nhưng xét modulo $20$:

$9^2=81\equiv1\pmod{20}$

$9^{50}\equiv1\pmod{20}$

Do đó:

$S=-\dfrac{9^{50}-1}{4}$

Ta có thể tính chính xác modulo $20$ bằng cách xét $9^{50}-1$ modulo $80$:

$9^2=81\equiv1\pmod{80}$

$9^{50}=(9^2)^{25}\equiv1\pmod{80}$

nên $S$ chia hết cho $20$.

Vậy: $S\vdots20$

Mà một số chia hết cho $20$ thì cũng chia hết cho $-20$.
b) Tính $S$

Từ: $4S=1-3^{100}$

Suy ra: $S=\dfrac{1-3^{100}}{4}$

Hay: $S=-\dfrac{3^{100}-1}{4}$

Từ đây: $4S=1-3^{100}$

Vì $4S$ chia hết cho $4$ nên:

$1-3^{100}\vdots4$

$3^{100}\equiv1\pmod4$

Vậy $3^{100}$ chia $4$ dư $1$.

Ta có:

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

\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)

\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\Rightarrowđpcm\)

Vậy \(A< \frac{3}{4}\)

10³⁰ > 2¹⁰⁰

9²⁰ > 27¹³

3²¹⁰ > 2³⁵⁰

\(m=1+3+3^2+...+3^{100}\)

\(3m=3+3^2+...+3^{101}\)

\(\Rightarrow3m-m=3^{101}-1\)

\(\Leftrightarrow m=\frac{3^{101}-1}{2}\)

\(M=1+3+3^2+...+3^{100}\)

\(\Rightarrow\)\(3M=3+3^2+3^3+...+3^{101}\)

        \(-M=1+3+3^2+...+3^{100}\)

\(\Rightarrow\)\(2M=3^{101}-1\)

\(\Rightarrow M=\frac{3^{101}-1}{2}\)

a) 3³ - 10² : 5² + 2³ . 7 

= 27 - 5⁴ : 5² + 8.7

= 27 - 5² + 56

= 27 - 25 + 56

= 2 + 56

= 58

a) 79

b) 2

c) 990

d) = 80  - {140 - [868-12(64)]}

    = 80 - (140-100)

    = 80- 40

    = 40

mk ko bít bạn có cần trình bày ko nên mk vít kq hoi

\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.5^3.2^4.6.7}=\frac{\left(-2\right)^2.3.3^2.5^3.7.8}{3.5^3.2\text{^2}.2^2.6.7}=\frac{3^2.8}{2^2.6}=\frac{9.8}{4.6}=\frac{3.3.4.2}{4.3.2}=3\)

Ta có

\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.5^3.2^4.42}=\frac{\left(-2\right)^3.3^2.7.8}{2^4.7.6}=\frac{-1.3^2.4}{2.3}=-1.3.2=-6\)

a, 100 + 98 + 96 + ... + 2 - 97 - 95 - 93 - ... - 1

= (100 + 98 + 96 + ... + 2) - (97 + 95 + 93 + ... + 1)

= 2550 - 2401

= 149

b, đặt A = 2 + 2² + 2³ + ... + 2¹⁰⁰

2A = 2² + 2³ + 2⁴ + ... + 2¹⁰¹

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

A = 2¹⁰¹ - 2

c, 3.3².3³....3¹⁰⁰

= 3^{1 + 2 + 3 + ... + 100}

= 3⁵⁰⁵⁰

bạn ơi câu c/ là phép nhân nhé

2¹ + 2² + 2³ + ... + 2¹⁰⁰

Ta có : S = 2 + 2² + 2³ + ... + 2¹⁰⁰

          2S = 2.(2 +  2² + 2³ + ... + 2¹⁰⁰)

          2S = 2² + 2³ + ... + 2¹⁰⁰ + 2¹⁰¹

        2S - S = (2² + 2³ + ... + 2¹⁰⁰ + 2¹⁰¹) - (2 +  2² + 2³ + ... + 2¹⁰⁰)

           S = 2¹⁰¹ - 2

\(2^1+2^2+2^3+...+2^{100}\)

Ta có : \(S=2+2^2+2^3+....2^{100}\)

         : \(2S=2.\left(2+2^2+2^3+....+2^{100}\right)\)

        : \(2S=2^2+2^3+.....+2^{100}+2^{101}\)

        : \(2S-S=\left(2^2+2^3+....+2^{100}+2^{101}\right)\)\(-\left(2+2^2+2^3+.....+2^{100}\right)\)

        : \(S=2^{101}-2\)