M = \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\)

=> 5M = 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)

=> 5M - M = ( 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)) - ( \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\))

4M = 1 - \(\left(\frac{1}{5}\right)^{50}\)

=> M = \(\frac{1-\left(\frac{1}{5}\right)^{50}}{4}\)< \(\frac{1}{4}\)

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

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) \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

= \(\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)\) - \(\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\) - \(\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)\) - 2.\(\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)\) - \(\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)

= \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\) - \(1-\frac{1}{2}-...-\frac{1}{100}\)

= \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

Vậy \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\) = \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

Mình chỉ làm được phần a) thôi, nhưng k cho mình nhé

\(B=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)

\(B=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)

\(B=\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+....+\frac{100-81}{81.100}\)

\(B=\frac{4}{1.4}-\frac{1}{1.4}+\frac{9}{4.9}-\frac{4}{4.9}+\frac{16}{9.16}-\frac{9}{9.16}+...+\frac{100}{81.100}-\frac{81}{81.100}\)

\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)

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

=> B < 1 (Đpcm)

B = 3/1².2² + 5/2².3² + 7/3².4² + ... + 19/9².10²

B = 3/1.4 + 5.4.9 + 7/9.16 + ... + 19/81.100

B = 1 - 1/4 + 1/4 - 1/9 + 1/9 - 1/16 + ... + 1/81 - 1/100

B = 1 - 1/100 < 1 ( đpcm)

_A=2^1+2^2+2^3+...+2^2010

A=(2^1+2^2)+(2^3+2^4)+...+(2^2009+2^2010)

A=2.(1+2)+2^3.(1+2)+...+2^2019.(1+2)

A=2.3+2^3.3+...+2^2009.3

A=3.(2+2^3+...+2^2009)

Vậy A chia hết cho 3.

_A=2^1+2^2+2^3+...+2^2010

A=(2^1+2^2+2^3)+(2^4+2^5+2^6)+...+ (2^2008+2^2009+2^2010)

A=2.(1+2+2^2)+2^4.(1+2+2^2)+...+2^2008.(1+2+2^2)

A=2.7+2^4.7+...+2^2008.7

A=7.(2+2^4+...+2^2008)

Vậy A chia hết cho 7.

=> A ⋮ 3, A ⋮ 7.

Lưu ý ^ là mũ nhé !!! (^-^)