ta có 

\(B=1+\left(1-\frac{1}{2}\right)+..+\left(1-\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=A\)

Vậy A=B

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

\(\Rightarrow\)\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)

\(\Rightarrow\)\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

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

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

\(\Rightarrow\)\(3B=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\)

\(\Rightarrow\)\(3B-B=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)

\(\Rightarrow\)\(2B=3-\frac{1}{3^{100}}\)

\(\Rightarrow\)\(B=\frac{3-\frac{1}{3^{100}}}{2}\)

c)

\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+....+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)

\(\left(1+1+1+....+1+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{6\times7}+\frac{1}{7\times8}\right)\)(Có  7 số 1)

\(7+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(7+1-\frac{1}{8}=\frac{63}{8}\)

Gợi ý 1 bài c) còn d) e) cũng làm như vậy nhé

Chúc bạn học tốt !!!

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

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

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

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

Tớ cx dg lm cái này :v

Bài 1 :

\(A=3^0+3^1+3^2+3^3+...+3^{98}\)

\(A=\left(1+3+3^2\right)+.....+\left(3^{97}+3^{98}+3^{99}\right)\) ( Nhóm 3 số 1 nhé )

\(A=13+.....+3^{97}.13⋮13\left(\text{đ}pcm\right)\)

Bài 2 :

Theo ý a ta có : 

\(A=13+.....+3^{97}.13+3^{99}+3^{100}\)

\(A=13+.....+3^{97}.13+3^{99}.4⋮̸13\)

Bài 3 :

Để D chia hết cho 2 thì x chia hết cho 2

1. \(A=3^0+3^1+3^2+...+3^{98}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\)

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

\(=13\left(1+3^3+...+3^{96}\right)\)chia hết cho \(13\).

2. \(B=3^0+3^1+3^2+3^3+...+3^{100}\)

\(=1+3+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=4+3^2\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)

\(=4+13\left(3^2+3^5+...+3^{98}\right)\)không chia hết cho \(13\).

3. \(D=\left(12.3+26.b+2022.c+x\right)\)chia hết cho \(2\)

\(\Leftrightarrow x⋮2\)(vì \(12.3⋮2,26b⋮2,2022c⋮2\))