A=1+2+2²+2³+...+2⁶²+2⁶³

2A=2+2²+2³+2⁴+...+2⁶³+2⁶⁴

2A-A=2+2²+2³+2⁴+...+2⁶³+2⁶⁴-1+2+2²+2³+...+2⁶²+2⁶³

A=2⁶⁴-1

\(A=1+2+2^2+2^3+..+2^{62}+2^{63}\)

\(2A=2+2^2+2^3+...+2^{63}+2^{64}\)

\(2A-A=2^{64}-1\)

\(A=2^{64}-1\)

A=1+2+2²+2³+...+2⁶²+2⁶³

2A=2(1+2+2²+2³+...+2⁶²+2⁶³)

2A=2+2³+2⁴+2⁵...+2⁶³+2⁶⁴

2A-A=(2+2³+2⁴+2⁵...+2⁶³+2⁶⁴)-(1+2+2²+2³+...+2⁶²+2⁶³)

A=2⁶⁴-1

Nha bạn. Chúc bn ht

A = 1 + 2 + 2² + 2³ + ... + 2⁶² + 2⁶³

2A = 2 + 2² + 2³ + 2⁴ + ... + 2⁶³ + 2⁶⁴

A = 2⁶⁴ - 1

\(S_1=1+2+2^2+2^3+..+2^{63}\\ \Rightarrow2S_1=2+2^2+2^3+2^4+...+2^{64}\\ \Rightarrow S_1-2S_1=1-2^{64}\\ \Rightarrow-S_1=1-2^{64}\\ \Rightarrow S_1=2^{64}-1.\)

- Ta có: S₁ = 1 + 2 + 2² + 2³ + … + 2⁶³ = 1 + 2(1 + 2 + 2² + 2³ + … + 2⁶²) (1)

= 1 + 2(S₁ - 2⁶³) = 1 + 2S₁ - 2⁶⁴ S₁ = 2⁶⁴ - 1

H2.right

`#3107.101107`

`S_1 = 1 + 2 + 2^2 + 2^3 + ... + 2^63`

`2S_1 = 2 + 2^2 + 2^3 + .... + 2^64`

`2S_1 - S_1 = (2 + 2^2 + 2^3 + ... + 2^64) - (1 + 2 + 2^2 + 2^3 + ... + 2^63)`

`S_1 = 2 + 2^2 + 2^3 + ... + 2^64 - 1 - 2 - 2^2 - 2^3 - ... - 2^63`

`S_1 = 2^64 - 1`

Vậy, `S_1 = 2^64 - 1.`

tính như nào nhỉ

`1+2+2^2+2^3+....+2^63`

`=2+2+2^2+2^3+....+2^63-1`

`=2.2+2^2+2^3+....+2^63-1`

`=2^2+2^2+2^3+....+2^63-1`

`=2.2^2+2^3+....+2^63-1`

`=2^3+2^3+...2^63-1`

`=2.2^3+....+2^63-1`

`=2^4+....+2^63-1`

`=2^{63}.2-1=2^64-1`

Đặt \(A=1+2+2^2+2^3+...+2^{63}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+2^5+...+2^{64}\)

\(\Rightarrow A=2A-A=\left(2+2^2+2^3+2^4+...+2^{64}\right)-\left(1+2+2^2+2^3+...+2^{63}\right)=2^{64}-1\left(đpcm\right)\)

\(=526+1590-262+795=2649\)

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

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

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

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

bài 1

2¹⁰¹ - 2

\(A=1+3+3^2+3^3+...+3^{99}\\ \Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)

\(\Rightarrow A=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=\left(1+3+3^2+3^3\right)\left(1+3^4+...+9^{96}\right)\)

\(\Rightarrow A=40\left(1+3^4+...+9^{96}\right)⋮40\)

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

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

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

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

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

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

2/3

Câu 1:

Program HOC24;

var i,n: integer;

t: longint;

write('Nhap N: '); readln(n);

t:=1;

for i:=1 to n do t:=t*i;

write(n,'! = ',t);

readln

end.

Câu 2:

Program HOC24;

var i,n: integer;

S: real;

begin

write('Nhap n: '); readln(n);

S:=0;

for i:=1 to n do S:=S+1/i;

write('S = ',s);

readln

end.

Nguyễn Quang Thành trả lời cụ thể đi