@Đỗ Nguyễn Như Bình \(\frac{2}{3^2}\) hay là \(\frac{2^2}{3}\) hay là \(\left(\frac{2}{3}\right)^2\) vậy em???????????

ta có : 1+1+1+1+1+1+1+1x0

=> 1x8 = 8

mà kòn x vs 0 nữa :

=> tổng đó =0

=> 0<3/4

=> E<3/4

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

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

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

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

\(6E=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6E-2E=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4E=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4E=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4E=3-\frac{203}{3^{100}}< 3\)

\(\Rightarrow4E< 3\)

\(\Rightarrow E< \frac{3}{4}\left(đpcm\right)\)

Bài 1:

Ta có: \(3+3^2+3^3+...+3^{100}\)

\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=120+3^5\left(3+3^2+3^3+3^4\right)+....+3^{96}\left(3+3^2+3^3+3^4\right)\)

\(=120+3^5.120+...+3^{96}.120\)

\(=120.\left(1+3^5+.....+3^{96}\right)\)

\(\Rightarrow3+3^2+3^3+3^4+....+3^{100}\)chia hết cho 120 (vì có chứa thừa số 120)

\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\)

\(\Leftrightarrow3E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\)

\(\Leftrightarrow3E-E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}-\frac{1}{3}-\frac{2}{3^2}-\frac{3}{3^3}-...-\frac{101}{3^{101}}\)

\(\Leftrightarrow2E=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{100}{3^{101}}\)

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

​​\(\Leftrightarrow3S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)​​

​\(\Leftrightarrow3S-S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)​

​\(\Leftrightarrow2S=1-\frac{1}{3^{100}}\)​

\(\Leftrightarrow S=\left(1-\frac{1}{3^{100}}\right)\div2\)

\(\Leftrightarrow2E=1+\left(1-\frac{1}{3^{100}}\right)\div2-\frac{101}{3^{101}}\)

​\(\Leftrightarrow2E=1+\frac{1}{2}-\frac{1}{3^{100}.2}-\frac{101}{3^{101}}\)​

​\(\Leftrightarrow2E=\frac{3}{2}-\frac{1}{3^{100}.2}-\frac{101}{3^{101}}< \frac{3}{2}\)​

\(\Leftrightarrow E< \frac{3}{4}\left(đpcm\right)\)

$A=\dfrac12-\dfrac{2}{2^2}+\dfrac{3}{2^3}-\dfrac{4}{2^4}+\cdots+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}$

Nhóm từng 2 số:

$A=\left(\dfrac12-\dfrac{2}{2^2}\right)+\left(\dfrac3{2^3}-\dfrac4{2^4}\right)+\cdots+\left(\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}\right)$

$=0+\dfrac18+\dfrac{2}{32}+\dfrac3{128}+\cdots+\dfrac{49}{2^{99}}$

$=\sum_{k=1}^{50}\dfrac{k-1}{2^{2k-1}}$

Ta có: $\dfrac{k-1}{2^{2k-1}}=\dfrac{2(k-1)}{4^k}$

Mà: $\sum_{k=1}^{\infty}\dfrac{k-1}{4^k}=\dfrac{1}{9}$

Nên: $A<2\cdot\dfrac19$ $=\dfrac29$

Vậy: $A<\dfrac29$

$4=1\cdot4,\quad28=4\cdot7,\quad70=7\cdot10,\ldots$

tức là: $E=\dfrac3{1\cdot4}+\dfrac3{4\cdot7}+\dfrac3{7\cdot10}+\cdots+\dfrac3{n(n+3)}$

Với $n=1,4,7,\ldots$.

Ta có: $\dfrac3{n(n+3)}=\dfrac1n-\dfrac1{n+3}$

Do đó: $E=\left(1-\dfrac14\right)+\left(\dfrac14-\dfrac17\right)+\left(\dfrac17-\dfrac1{10}\right)+\cdots+\left(\dfrac1n-\dfrac1{n+3}\right)$

$=1-\dfrac1{n+3}$

Vì: $\dfrac1{n+3}>0$ nên: $1-\dfrac1{n+3}<1$

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

\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)

\(\Leftrightarrow2A=3^{101}-3\)

\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)

\(\Leftrightarrow A< B\)

a. tính A = 3+3^2+3^3+3^4+.....+3^100

3A=3^2+3^3+3^4+3^5+....+3^100

3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100

mà B=3^100-1 => A<B