a) \(S=1+2+2^2+2^3+...+2^{2022}=\dfrac{2^{2022+1}-1}{2-1}=2^{2023}-1\)

b) \(S=1+4+4^2+4^3+...+4^{2022}=\dfrac{4^{2022+1}-1}{4-1}=\dfrac{4^{2023}-1}{3}\)

\(S=1+2+2^2+2^3+...+2^{2022}\\ 2S=2+2^2+2^3+2^4+...+2^{2023}\\ 2S-S=2+2^2+2^3+2^4+...+2^{2023}-1-2-2^2-2^3-...-2^{2022}\\ S=2^{2023}-1\\ S=4+4^2+4^3+...+4^{2022}\\ 4S=4^2+4^3+4^4+...+4^{2023}\\ 4S-S=4^2+4^3+4^4+...+4^{2023}-4-4^2-4^3-...-4^{2023}\\ 3S=4^{2023}-4\\ S=\dfrac{4^{2023}-4}{3}\)

Bài 1

a) S = 1 + 2 + 2² + 2³ + ... + 2²⁰²³

2S = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²⁴

S = 2S - S = (2 + 2² + 2³ + ... + 2²⁰²⁴) - (1 + 2 + 2² + 2³)

= 2²⁰²⁴ - 1

b) B = 2²⁰²⁴

B - 1 = 2²⁰²⁴ - 1 = S

B = S + 1

Vậy B > S

a,

\(S=1+2+2^2+...+2^{2023}\)

\(2S=2+2^2+2^3+...+2^{2024}\)

\(\Rightarrow S=2^{2024}-1\)

b.

Do \(2^{2024}-1< 2^{2024}\)

\(\Rightarrow S< B\)

2.

\(H=3+3^2+...+3^{2022}\)

\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)

\(\Rightarrow3H-H=3^{2023}-3\)

\(\Rightarrow2H=3^{2023}-3\)

\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)

\(A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)=3+2^2.3+...+2^{10}.3=3\left(1+2^2+...+2^{10}\right)⋮3\)

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

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

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

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

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

\(\Rightarrow2B=2^2+2^3+2^4+...+2^{101}\)

\(\Rightarrow2B-B=2^2+2^3+2^4+...+2^{101}-2-2^2-2^3-...-2^{100}\)

\(\Rightarrow2B-B=2^{101}-2\)

bài 1

2¹⁰¹ - 2

Lời giải:
$E=1-2+22-23+24-25+.....+21000$

$=(1-2)+(22-23)+(24-25)+......+(20998-20999)+21000$
$=(-1)+(-1)+(-1)+....+(-1)+21000$

Số lần xuất hiện của -1: $[(20999-22):1+1]:2+1=10490$

$E=(-1).10490+21000=10510$

bít kq nhưng ko thích giải

cậu ko giúp cậu ấy thì thôi đừng bảo như thế

a.

$S=1+2+2^2+2^3+...+2^{2017}$
$2S=2+2^2+2^3+2^4+...+2^{2018}$

$\Rightarrow 2S-S=(2+2^2+2^3+2^4+...+2^{2018}) - (1+2+2^2+2^3+...+2^{2017})$

$\Rightarrow S=2^{2018}-1$

b.

$S=3+3^2+3^3+...+3^{2017}$
$3S=3^2+3^3+3^4+...+3^{2018}$

$\Rightarrow 3S-S=(3^2+3^3+3^4+...+3^{2018})-(3+3^2+3^3+...+3^{2017})$

$\Rightarrow 2S=3^{2018}-3$
$\Rightarrow S=\frac{3^{2018}-3}{2}$
 

Câu c, d bạn làm tương tự a,b. 

c. Nhân S với 4. Kết quả: $S=\frac{4^{2018}-4}{3}$

d. Nhân S với 5. Kết quả: $S=\frac{5^{2018}-5}{4}$

\(\dfrac{2}{3}\)

=\(\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{1}{22}-\dfrac{23}{22}\right)+\dfrac{2}{3}\)

=\(\dfrac{17}{17}-\dfrac{22}{22}+\dfrac{2}{3}\)

=\(1-1+\dfrac{2}{3}\)

=0+\(\dfrac{2}{3}\)

=\(\dfrac{2}{3}\)