Vì tui dùng app giải

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2021}}\)

=>\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2020}}\)

=>\(3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2020}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{2021}}\)

=>\(2A=1-\dfrac{1}{3^{2021}}\)

=>\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{2021}}< \dfrac{1}{2}\)

help me pls

a) Gọi ƯCLN(12n+1;30n+2) = d

\(\Rightarrow\begin{cases}12n+1⋮d\\30n+2⋮d\end{cases}\)

\(\Rightarrow\begin{cases}5\left(12n+1\right)⋮d\\2\left(30n+2\right)⋮d\end{cases}\)

\(\Rightarrow\begin{cases}60n+5⋮d\\60n+4⋮d\end{cases}\)

=> ( 60n + 5 ) - ( 60n + 4 ) \(⋮\) d

=> 1 \(⋮\) d

=> d = 1

Vậy \(\frac{12n+1}{30n+2}\) là phân số tối giản

b) Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

                \(\frac{1}{3^2}< \frac{1}{2.3}\)

                 .........

                  \(\frac{1}{100^2}< \frac{1}{99.100}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\) 

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}< 1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\) ( đpcm )

help me

Ta có :

\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+...+\left(1-\frac{99}{100}\right)\right]\)

\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)

\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)

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

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

ta có 100-(1+1/2+1/3+.....+1/100)

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

=(1-1)+(1-1/2)+(1-1/3)+.......+(1-1/100)

=1/2+2/3+3/4+.....+99/100

mẹ mày béo

@Phí Thị Thanh Duyên không bình luận xúc phạm nha bạn.