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

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: $1-\dfrac1{n+3}<1$

E = 3 / 4+ 3 / 28 +......+ 3 / n . ( n + 3 )

E = 3 / 1 . 4 + 3 / 4 . 7 +...+ 3 / n ( n + 3 )

E = 1 -1/ 4 + 1 / 4 - 1 /7 +......+ 1 / n - 1 / n + 3

E = 1 - 1 / n + 3

E = n + 2 / n + 3

đề bài tìm n hay tìm E z?

b) D = \(\frac{3}{4}+\frac{3}{8}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)

    D =  \(3\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)

    D = \(3\left(\frac{1}{1x4}+\frac{1}{4x7}+\frac{1}{7x10}+\frac{1}{10x13}+\frac{1}{13x16}+\frac{1}{16x19}\right)\)

    D = \(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)

Chắc vậy

3/4+3/28+....+3/n.(n+3)=3/1.4+3/4.7+....+3/n.(n+3)=1/1-1/4+1/4-1/7+...+1/n-1/n+3=1-1/n+3.

Suy ra E<1

\(E=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)

\(\Rightarrow E=1+\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+\left(-\frac{1}{10}+\frac{1}{10}\right)+...\left(-\frac{1}{n}+\frac{1}{n}\right)-\frac{1}{n+3}\)

\(E=1-\frac{1}{n+3}<1\)       (ĐPCM)

Có trong Violympic Toán vòng 3 đó đáp án là 719 nha !!!

a; 1, 3 ,6, 10, 15...

St1 = 1 = 1.2 : 2

St2 = 3 = 2.3: 2

St3 = 6 = 3.4 : 2

St4 = 10 = 4.5 : 2

St5 = 15 = 5.6 : 2

Stn = n.(n+1) : 2

b) 4, 28, 70, 130, 208...

Stn = 4 = 1.4 = (3.1 - 2).(3.1 + 1)

St2 = 28 = 4.7 = (3.2 - 2).(3.2 + 1)

St3 = 70 = 7.10 = (3.3 - 2).(3.3 + 1)

St4 = 130 = 10.13 = (3.4 - 2).(3.4 + 1)

St5 = 208 = 13.16 = (3.5 - 2).(3.5 + 1)

STn = (3n - 2).(3n + 1)

B = \(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)

= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\)

= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{13}-\frac{1}{16}\)

= \(1-\frac{1}{16}\)

= \(\frac{15}{16}\)