nhận xét :

\(\frac{1}{2^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{3^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

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

\(\frac{1}{100^2}=\frac{1}{100.101}=\frac{1}{100}-\frac{1}{101}\)

vậy

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{101}=\frac{9}{202}< \frac{3}{4}\)

Ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};.....;\frac{1}{100^2}< \frac{1}{99.100}\)

=>\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

=>\(S< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

=>\(S< \frac{1}{4}+\frac{1}{2}-\frac{1}{100}=\frac{3}{4}-\frac{1}{100}< \frac{3}{4}\)

=>S<3/4(đpcm)

ta có

1/3^2 < 1/2*3 ; 1/4^2 < 1/3*4 ; .........; 1/100^2< 1/99*100

suy ra s=1/2^2+1/3^2+....+1/100^2 < 1/2*3 + 1/3*4 +...........+ 1/99*100

S < 1/4 + 1/2 - 1/3 + 1/3 +..........+ 1/99 - 1/100

suy ra S< 1/4 +1/2 - 1/100

hay S < 3/4 -1/100

mà 3/4 -1/100< 3/4

suy ra s<3/4

Ta có 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

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

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)

ĐPcm

Giải:

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

Ta có:

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

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

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\) 

\(...\) 

\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\) 

\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\) 

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

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

\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\) 

\(\Rightarrow S< 1\) 

Vậy S < 1.

\(S=1+3^2+3^4+...+3^{100}\)

\(\Rightarrow9S=3^2+3^4+....+3^{102}\)

\(\Rightarrow9S-S=\left(3^2+....+3^{102}\right)-\left(1+....+3^{100}\right)\)

\(\Rightarrow8S=3^{102}-1=9^{51}-1>8^{51}:2=2^{152}\)

bạn ghi sai đề rồi . đáng là 1+3^1 trước chứ

\(3s=3-3^2+3^3-3^4+...+3^{100}\)

\(4s=\left(3-3^2+3^3-3^4+...+3^{101}\right)+\left(1-3+3^2-3^3+...+3^{100}\right)\)

\(4s=1\)

\(s=\dfrac{1}{4}>\dfrac{1}{5}\)