Ta có : \(\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}\)

= \((1-\frac{1}{2})+(1-\frac{1}{3})+...+(1-\frac{99}{100})\)(100 cặp số )

= \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)(100 số hạng 1)

= \(1\times100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{100}\right)\)

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

=> 100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100

Bạn cố giải cho mình dễ hiểu hơn ko?

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

\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{2}{5}\)

nhưng tại sao lại >1/2*3+1/3*4+1/4*5+...+1/9*10

bạn giúp mk nha

Nhân Q cho 3 ói lấy 3Q-Q sẽ ra 2Q=? =>Q òi so sánh

Đặt A=\(2+2^2+2^3+...+2^{99}\)
2A=\(2^2+2^3+2^4+...+2^{100}\)
=>2A-A=(\(2^2+2^3+2^4+...+2^{100}\))-(\(2+2^2+2^3+...+2^{99}\))
=>A=\(2^{100}-2\)
Thay vào đề bài :
\(1+2^{100}-2=2^{100}-1\)
=>\(\left(1-2\right)+2^{100}=2^{100}-1\)
=>

=>\(-1+2^{100}=2^{100}-1\)
=>\(2^{100}-1=2^{100}-1\left(đpcm\right)\)

1/2+1/3+1/4+...+1/63>1/31+1/31+...+1/31(62 số hạng 1/31)

hay 1/2+1/3+1/4+...+1/63>62 x 1/31

nên 1/2+1/3+1/4+...+1/63>2(dpcm)

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\) \(\Rightarrow A< \dfrac{99}{100}\)

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

a: Ta có

A = \(\dfrac{1}{10}\) + \((\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + ...+ \(\dfrac{1}{100}\)\()\)

⇒ A > \(\dfrac{1}{10}\) + \((\dfrac{1}{100}\) + \(\dfrac{1}{100}\) + ...+ \(\dfrac{1}{100}\)\()\)90 số hạng 

⇒ A > \(\dfrac{1}{10}\) + \(\dfrac{90}{100}\)

⇒ A > 1

vậy A > 1

b: ta có

S = (\(\dfrac{1}{21}\) + \(\dfrac{1}{22}\)+ \(\dfrac{1}{23}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{26}\) + \(\dfrac{1}{27}\)+ \(\dfrac{1}{28}\) + \(\dfrac{1}{29}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{33}\) + \(\dfrac{1}{34}\) + \(\dfrac{1}{35}\))

⇒ S > (\(\dfrac{1}{25}\) + \(\dfrac{1}{25}\)+ \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{30}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{35}\) + \(\dfrac{1}{35}\)+ \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\))

⇔ S > \(\dfrac{5}{25}\)+\(\dfrac{5}{30}\)+\(\dfrac{5}{35}\)

⇔ S > \(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)

⇔ S > \(\dfrac{107}{210}\)> \(\dfrac{105}{210}\)=\(\dfrac{1}{2}\)

vậy S > \(\dfrac{1}{2}\)