a) \(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)

b) \(B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+...+3^{101}\)

\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)

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

c) \(C=5+5^2+...+5^{30}\)

\(\Rightarrow5C=5^2+5^3+...+5^{31}\)

\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)

\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)

d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)

\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)

\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)

1990.1990 -1992.1988

\(A=2+2^2+...+2^{20}\)

\(2A=2^2+2^3+...+2^{21}\)

\(2A-A=2^2+2^3+...+2^{21}-2-2^2-...-2^{20}\)

\(A=2^{21}-2\)

___________

\(B=5+5^2+...+5^{50}\)

\(5B=5^2+5^3+...+5^{51}\)

\(5B-B=5^2+5^3+...+5^{51}-5-5^2-...-5^{50}\)

\(4B=5^{51}-5\)

\(B=\dfrac{5^{51}-5}{4}\)

___________

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

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

\(3C-C=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}\)

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

\(C=\dfrac{3^{101}-1}{2}\)

2A= 2(2+2²+2³+...+2¹⁹+2²⁰)

2A= 2²+2³+2⁴+...+2²⁰+2²¹

2A-A=(2²+2³+2⁴+...+2²⁰+2²¹)-(2+2²+2³+...+2¹⁹+2²⁰)

A=2²¹-2

Vậy A=2²¹-2

Làm tương tự nhee

khó v

Bài 1:

1) \(9A=3^3+3^5+...+3^{113}\)

\(\Rightarrow8A=9A-A=3^3+3^5+...+3^{113}-3-3^3-...-3^{111}=3^{113}-3\)

\(\Rightarrow A=\dfrac{3^{113}-3}{8}\)

2) \(9B=3^4+3^6+...+3^{202}\)

\(\Rightarrow8B=9B-B=3^4+3^6+...+3^{202}-3^2-3^4-...-3^{200}=3^{202}-3^2=3^{202}-9\)

\(\Rightarrow B=\dfrac{3^{202}-9}{8}\)

3) \(25C=5^3+5^5+...+5^{101}\)

\(\Rightarrow24C=25C-C=5^3+5^5+...+5^{101}-5-5^3-...-5^{99}=5^{101}-5\)

\(\Rightarrow C=\dfrac{5^{101}-5}{24}\)

4) \(25D=5^4+5^6+...+5^{102}\)

\(\Rightarrow24D=25D-D=5^4+5^6+...+5^{102}-5^2-5^4-...-5^{100}=5^{102}-25\)

\(\Rightarrow D=\dfrac{5^{102}-25}{24}\)

Bài 2:

a) Gọi d là UCLN(2n+1,n+1)

\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\n+1⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\2n+2⋮d\end{matrix}\right.\)

\(\Rightarrow\left(2n+2\right)-\left(2n+1\right)⋮d\Rightarrow1⋮d\)

Vậy 2n+1 và n+1 là 2 số nguyên tố cùng nhau

\(\Rightarrow\dfrac{2n+1}{n+1}\) là phân số tối giản

b) Gọi d là UCLN(2n+3,3n+4)

\(\Rightarrow\left\{{}\begin{matrix}2n+3⋮d\\3n+4⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}6n+9⋮d\\6n+8⋮d\end{matrix}\right.\)

\(\Rightarrow\left(6n+9\right)-\left(6n+8\right)⋮d\Rightarrow1⋮d\)

\(\Rightarrow\dfrac{2n+3}{3n+4}\) là phân số tối giản

Câu 3:

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

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

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

\(2A=3^{101}-3\) 

Mà: \(2A+3=3^N\)

\(\Rightarrow3^{101}-3+3=3^N\)

\(\Rightarrow3^{101}=3^N\)

\(\Rightarrow N=101\)

Vậy: ... 

Câu 1:

\(A=4+2^2+...+2^{20}\)

Đặt \(B=2^2+2^3+...+2^{20}\)

=>\(2B=2^3+2^4+...+2^{21}\)

=>\(2B-B=2^3+2^4+...+2^{21}-2^2-2^3-...-2^{20}\)

=>\(B=2^{21}-4\)

=>\(A=B+4=2^{21}-4+4=2^{21}\) là lũy thừa của 2

Câu 6:

Đặt A=1+2+3+...+n

Số số hạng là \(\dfrac{n-1}{1}+1=n-1+1=n\left(số\right)\)

=>\(A=\dfrac{n\left(n+1\right)}{2}\)

=>\(A⋮n+1\)

Câu 5:

\(A=5+5^2+...+5^8\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+\left(5^7+5^8\right)\)

\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+5^6\left(5+5^2\right)\)

\(=30\left(1+5^2+5^4+5^6\right)⋮30\)

a) \(S=1+2+2^2+..+2^{2022}\)

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

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

\(S=2^{2023}-1\)

b) \(S=3+3^2+3^3+...+3^{2022}\)

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

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

\(2S=3^{2023}-3\)

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

c) \(S=4+4^2+4^3+...+4^{2022}\)

\(4S=4^2+4^3+...+4^{2023}\)

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

\(3S=4^{2023}-4\)

\(S=\dfrac{4^{2023}-4}{3}\)

d) \(S=5+5^2+...+5^{2022}\)

\(5S=5^2+5^3+...+5^{2023}\)

\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)

\(4S=5^{2023}-5\)

\(S=\dfrac{5^{2023}-5}{4}\)

siuu

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: $S=\frac{5^{2018}-5}{4}$

 a)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)

b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)

c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)

\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)

\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)

Tương tự câu d,e,f bạn tự làm nhé

mn mn ơiii

helllppppppppp

\(2,\\ 3^{n-3}+2^{n-3}+3^{n+1}+2^{n+2}\\ =3^{n-3}\left(1+3^4\right)+2^{n-3}\left(1+2^5\right)\\ =3^{n-3}\cdot82+2^{n-3}\cdot33\)

Vì \(3^{n-3}\cdot82⋮2;⋮3\) nên \(3^{n-3}\cdot82⋮6\)

\(2^{n-3}\cdot33⋮2;⋮3\) nên \(2^{n-3}\cdot33⋮6\)

Do đó tổng trên chia hết cho 6 với mọi \(n\in N\)

a: =>3[(2x-1)^2-4]=49*125:175+196=231

=>(2x-1)^2-4=77

=>(2x-1)^2=81

=>2x-1=9 hoặc 2x-1=-9

=>x=5 hoặc x=-4

b: \(\Leftrightarrow2\cdot3^x\cdot3-4^3=7^2\cdot\left(27-25\right)\)

=>\(6\cdot3^x=49\cdot2+64=162\)

=>3^x=27

=>x=3