Giải:

A=1/2²+1/3²+1/4²+...+1/9²

Ta có:

1/2²<1/1.2

1/3²<1/2.3

1/4²<1/3.4

...

1/9²<1/8.9

⇒A<1/1.2+1/2.3+1/3.4+...+1/8.9

A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9

A<1/1-1/9

A<8/9

Ta có:

1/2²>1/2.3

1/3²>1/3.4

1/4²>1/4.5

...

1/9²>1/9.10

⇒A>1/2.3+1/3.4+1/4.5+...+1/9.10

A>1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10

A>1/2-1/10

A>2/5

Vậy 2/5<A<8/9 (đpcm)

Chúc bạn học tốt!

Ta thấy:

\(2^2=2.2>1.2\Rightarrow\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(3^2=3.3>2.3\Rightarrow\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

.................

\(9^2=9.9>8.9\Rightarrow\dfrac{1}{9^2}< \dfrac{1}{8.9}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)

\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\)

=> Đpcm

Ta thấy:

22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2

32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3

.................

92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9

⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9

⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89

=> ...(tự viết)

Ta thấy:

22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2

32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3

.................

92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9

⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9

⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89

=> 11111111111111111111110101010110000

HACK

cho sửa lại là:

Ta thấy:

\(⇒ \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... < \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

\(⇔\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2} <1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4} ...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)

⇒ {{{tự làm nhé}}}

Tính các tổng sau:

1, S=1-2+3_4+..+25-26

S =-1+3-5+7-...-53+55                       ( có 28 số hạng )

   = (-1+3)+(-5+7)+...+(-53+55)         ( có 28:2=14 nhóm )

   = 2+2+...+2

    = 2 . 14

     = 28

Á nhầm rùi xl bn nha

3 

S =-1+3-5+7-...-53+55     ( có 28 số hạng )

   = (-1+3)+(-5+7)+...+(-53+55)      ( có 28:2=14 nhóm )

   = 2+2+...+2

    = 2 . 14

     = 28

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

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

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

a) \(S=1+2+2^2+..+2^{2022}\)

\(2S=2+2^2+2^3+...+2^{2023}\)

\(2S-S=2+2^2+2^3+...+2^{2023}-1-2-2^2-...-2^{2022}\)

\(S=2^{2023}-1\)

b) \(S=3+3^2+3^3+...+3^{2022}\)

\(3S=3^2+3^3+...+3^{2023}\)

\(3S-S=3^2+3^3+....+3^{2023}-3-3^2-...-3^{2022}\)

\(2S=3^{2023}-3\)

\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)

c) \(S=4+4^2+4^3+...+4^{2022}\)

\(4S=4^2+4^3+...+4^{2023}\)

\(4S-S=4^2+4^3+...+4^{2023}-4-4^2-...-4^{2022}\)

\(3S=4^{2023}-4\)

\(S=\dfrac{4^{2023}-4}{3}\)

d) \(S=5+5^2+...+5^{2022}\)

\(5S=5^2+5^3+...+5^{2023}\)

\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)

\(4S=5^{2023}-5\)

\(S=\dfrac{5^{2023}-5}{4}\)

1: \(=\dfrac{1}{29\cdot30}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{28\cdot29}\right)\)

\(=\dfrac{1}{29\cdot30}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{28}-\dfrac{1}{29}\right)\)

\(=\dfrac{1}{29\cdot30}-\dfrac{28}{29}=\dfrac{1-28\cdot30}{870}=\dfrac{-859}{870}\)

a.

$S=1+2+2^2+2^3+...+2^{2017}$
$2S=2+2^2+2^3+2^4+...+2^{2018}$

$\Rightarrow 2S-S=(2+2^2+2^3+2^4+...+2^{2018}) - (1+2+2^2+2^3+...+2^{2017})$

$\Rightarrow S=2^{2018}-1$

b.

$S=3+3^2+3^3+...+3^{2017}$
$3S=3^2+3^3+3^4+...+3^{2018}$

$\Rightarrow 3S-S=(3^2+3^3+3^4+...+3^{2018})-(3+3^2+3^3+...+3^{2017})$

$\Rightarrow 2S=3^{2018}-3$
$\Rightarrow S=\frac{3^{2018}-3}{2}$
 

Câu c, d bạn làm tương tự a,b. 

c. Nhân S với 4. Kết quả: $S=\frac{4^{2018}-4}{3}$

d. Nhân S với 5. Kết quả: $S=\frac{5^{2018}-5}{4}$

Thanks trước

helppp me?

Lời giải:

$S=\frac{1}{2^2}+\frac{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}$

$> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}$
$=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}$
$=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}(*)$

Lại có:

$S=\frac{1}{2^2}+\frac{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}$

$< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}(**)$
Từ $(*); (**)$ ta có đpcm.