a: Ta có công thức tổng quát:

\(1-\frac{1}{1+2+\cdots+n}\)

\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}=1-\frac{2}{n\left(n+1\right)}\)

\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Ta có: \(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+\cdots+2022}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(2022+2\right)\left(2022-1\right)}{2022\left(2022+1\right)}\)

\(=\frac{4\cdot5\cdot\ldots\cdot2024}{3\cdot4\cdot\ldots\cdot2023}\cdot\frac{1\cdot2\cdot\ldots\cdot2021}{2\cdot3\cdot\ldots\cdot2022}=\frac{2024}{3}\cdot\frac{1}{2022}=\frac{1012}{1011\cdot3}=\frac{1012}{3033}\)

b:Sửa đề: \(B=1+\frac12\left(1+2\right)+\frac13\left(1+2+3\right)+\cdots+\frac{1}{100}\left(1+2+\cdots+100\right)\)

\(=1+\frac12\cdot\frac{2\cdot3}{2}+\frac13\cdot\frac{3\cdot4}{2}+\cdots+\frac{1}{100}\cdot\frac{100\cdot101}{2}\)

\(=1+\frac32+\frac42+\cdots+\frac{101}{2}=\frac12\left(2+3+4+\cdots+101\right)\)

\(=\frac12\left(101-2+1\right)\cdot\frac{101+2}{2}=\frac12\cdot100\cdot\frac{101+2}{2}=103\cdot25=2575\)

TA có: \(3a-b=\frac12\left(a+b\right)\)

=>2(3a-b)=a+b

=>6a-2b=a+b

=>5a=3b

=>a=0,6b

\(C=\frac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)

\(=\frac{\left(0,6b\right)^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)

\(=\frac{\left(0,6\right)^{2022}\left(b^{2022}+5^{2022}\right)}{b^{2022}+5^{2022}}=\left(0,6\right)^{2022}=\left(\frac35\right)^{2022}\)

B=13-5+2022=2030

\(\left|x-1\right|+\left(y+2\right)^{2022}=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\\left(y+2\right)^{2022}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\ \Rightarrow B=13.1-5\left(-8\right)+2022=13+40+2022=2075\)

=A*(1/100-1/10^2)

=0

1/

$A=1+2-3-4+5+6-7-8+....+2017+2018-2019-2020+2021+2022$

$=(1+2-3-4)+(5+6-7-8)+...+(2017+2018-2019-2020)+4043$
$=(-4)+(-4)+(-4)+...+(-4)+4043$
Số lần xuất hiện của -4 là: $[(2020-1):1+1]:4=505$

$A=(-4)\times 505+4043=2023$

Câu b có vẻ đề sai. Bạn xem lại nhé.

Bài 3:

a: a*S=a^2+a^3+...+a^2023

=>(a-1)*S=a^2023-a

=>\(S=\dfrac{a^{2023}-a}{a-1}\)

b: a*B=a^2-a^3+...-a^2023

=>(a+1)B=a-a^2023

=>\(B=\dfrac{a-a^{2023}}{a+1}\)

|x - 2| + |y - 1| + (x - y - z)²⁰²² = 0 (1)

Do |x - 2| ≥ 0 với mọi x ∈ R

|y - 1| ≥ 0 với mọi x ∈ R

(x - y - z)²⁰²² ≥ 0 với mọi x ∈ R

(1) ⇒  |x - 2| = |y - 1| = (x - y - z)²⁰²² = 0

*) |x - 2| = 0

x - 2 = 0

x = 2

*) |y - 1| = 0

y - 1 = 0

y = 1

*) (x - y - z)²⁰²² = 0

x - y - z = 0

2 - 1 - z = 0

1 - z = 0

z = 1

⇒ C = 26x - 3y²⁰²² + z²⁰²³

= 26.2 - 3.1²⁰²² + 1²⁰²³

= 52 - 3 + 1

= 50

Lời giải:
Đặt $2021=a$ thì:
$A=a^2+(a+1)^2+(a+2)^2+(a+3)^2$
$=4a^2+12a+14=(2a+3)^2+5=4045^2+5$ chia hết cho $25$ nhưng không chia hết cho $5$

Do đó $A$ không là số chính phương 

-----------------------

$9\equiv 1\pmod 4\Rightarrow 9^{100}\equiv 1\pmod 4$

$94^{100}\equiv 0\pmod 4$

$1994^{100}\equiv 0\pmod 4$

$\Rightarrow B\equiv 1+1+0+1\equiv 2\pmod 4$

Một scp không thể chia 4 dư 2 nên $B$ không là scp

---------------

Công thức $1^3+2^3+...+n^3=[\frac{n(n+1)}{2}]^2$ là scp nên $C$ là scp.

A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^100

2A = 1 + 1/2 + 1/2^2 + ... + 1/2^99
2A - A = (1 + 1/2 + 1/2^2 + ... + 1/2^99) - (1/2 + 1/2^2 + 1/2^3 + ... + 1/2^100)

A = 1 - 1/2^100

B = 1 + 1/3 + 1/3^3 + ... + 1/3^2022

3B = 3 + 1 + 1/3 + ... + 1/3^2021

3B - B = (3 + 1 + 1/3 + ... + 1/3^2021) - (1 + 1/3 + 1/3^3 + ... + 1/3^2022)

2B = 3 - 1/3^2022

B = \(\dfrac{\text{3 - 1/3^2022}}{\text{2}}\)

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

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

2A -A   =  1 - \(\dfrac{1}{2^{100}}\)

       A  =   1 - \(\dfrac{1}{2^{100}}\)

B          =    1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^3}\) + ....+ \(\dfrac{1}{3^{2022}}\) 

Xem lại đề bài