a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)

... . . . .

\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)

\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)

b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

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

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

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

Suy ra \(\frac{2}{5}< S\) (1)

Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

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

Từ đó suy ra S < 8/9

Từ (1) và (2) suy ra đpcm

Mik lười quá bạn tham khảo câu 3 tại đây nhé:

Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath

\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)

\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)

\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)

\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)

\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}.\)

\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}-\frac{1}{4}-\frac{2}{4^2}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)

\(3S< A=1+\frac{1}{4}+...+\frac{1}{4^{2018}}\)\(\Rightarrow3A=4A-A=4-\frac{1}{4^{2018}}< 4\)(sau khi rút gọn)

\(\Rightarrow3.3S< 4\Rightarrow9S< 4\)

\(\Rightarrow S< \frac{4}{9}< \frac{1}{2}\)

Ta có : S =\(\frac{1}{2^2}\)\(+\)\(\frac{1}{3^2}\)\(+\)...\(+\)\(\frac{1}{9^2}\)

              = \(\frac{1}{2.2}\)\(+\)\(\frac{1}{3.3}\)\(+\)...\(+\)\(\frac{1}{9^2}\)

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

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

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

\(\Rightarrow\)S <  \(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)...\(+\)\(\frac{1}{8.9}\)

            =\(1\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)...\(+\)\(\frac{1}{8}\)\(-\)\(\frac{1}{9}\)

            = \(1\)\(-\)\(\frac{1}{9}\)= \(\frac{8}{9}\)

Vậy \(\frac{2}{5}\)< S < \(\frac{8}{9}\)(đpcm)

Chúc bạn học tốt

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

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

\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(S>\frac{1}{2}-\frac{1}{10}\)

\(S>\frac{4}{10}=\frac{2}{5}\)

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

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9.10}< S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{8.9}\)

=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}< S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{8}-\frac{1}{9}\)

=> \(\frac{1}{2}-\frac{1}{10}< S< 1-\frac{1}{9}\)

=> \(\frac{2}{5}< S< \frac{8}{9}\)(dpcm )

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)

Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ;   (1/51 + 1/52+...+1/59+1/60) > 1/6

S > 1/4 + 1/5 + 1/6.

Trong khi đó (1/4 + 1/5 + 1/6) > 3/5

Vậy S > 3/5

Phần 2. 

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)

Tương tự : (1/41 + 1/42 + ...+ 1/50)  < 1/4 ;   (1/51 + 1/52+...+1/59+1/60) < 1/5

Mà S = (1/3 + 1/4 + 1/5) < 4/5 (Vì 1/3 + 1/5 < 3/5 hay 7/12 < 3/5 hay 35/60 < 36/60)

Vậy S <  4/5

Vậy \(\frac{3}{5}< S< \frac{4}{5}\)