Sửa đề: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{100^2}<\frac34\)

Ta có: \(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

\(\frac{1}{4^2}<\frac{1}{3\cdot4}=\frac13-\frac14\)

...

\(\frac{1}{100^2}<\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)

Do đó: \(\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{100^2}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}=\frac12-\frac{1}{100}<\frac12\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{100^2}<\frac14+\frac12=\frac34\)

\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(B=\dfrac{1}{2.2}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)

\(B=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{100}\)

\(B=0+0+...+0\)

\(B=0\)

B=221​+421​+621​+...+10021​

�=12.2+14.4+...+1100.100B=2.21​+4.41​+...+100.1001​

�=12−12+14−14+...+1100−1100B=21​−21​+41​−41​+...+1001​−1001​

$B=0+0+...+0$

$B=0$

$tick\; cái$

bạn sao chép của người bạn Ninh mà

a:

Số số hạng trong dãy M là:

(1002-12):10+1=100(số)

=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10

\(M=1002-992+982-972+...+22-12\)

\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)

\(=10+10+...+10\)

=10*50=500

b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)

\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)

=10+10+...+10

=10*10=100

fan bé sans à

wuttttt

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

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

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

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

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

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

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

ủa bạn ơi, lớn hơn 1/2 hay bé hơn 1/2 vậy bạn

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}}}