gọi biểu thức cần CM là B

\(3B=1+\frac23+\frac{3}{2^2}+\frac{4}{3^3}+\cdots+\frac{100}{3^{99}}\)

=> \(3B-B=1+\left(\frac23-\frac13\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+\cdots+\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)-\frac{100}{3^{100}}\)

\(2B=1+\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

đặt C= \(\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{99}}\)

=> \(3C=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\)

=> \(3C-C=\left(1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\right)-\left(\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}\right)\)

\(2C=1-\frac{1}{3^{99}}\)

=> \(C=\frac12-\frac{1}{2\cdot3^{99}}\)

\(2B=1+\frac12-\left(\frac{1}{2\cdot3^{99}}+\frac{100}{3^{100}}\right)\)

vì trong ngoặc lớn hơn 0

=> \(2B<\frac32\)

\(B<\frac34\left(đpcm\right)\)

\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\left(\frac{1}{4\cdot5}+.\ldots+\frac{1}{99\cdot100}\right)\)

\(A=\frac{7}{12}+\left(\frac{1}{4\cdot5}+\cdots+\frac{1}{99\cdot100}\right)\)

Mà \(\left(\frac{1}{4\cdot5}+\cdots+\frac{1}{99\cdot100}\right)>0\)

=> A>\(\frac{7}{12}\)

mặt khác ta có: \(A=1-\frac12+\frac13-\frac14+\frac14-\frac15+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\left(\frac12-\frac13\right)-\left(\frac14-\frac15\right)-.\ldots-\left(\frac{1}{98}-\frac{1}{98}\right)-\frac{1}{100}\)

\(A=\frac56-\left(\frac14-\frac15\right)-.\ldots-\left(\frac{1}{98}-\frac{1}{98}\right)-\frac{1}{100}\)

=> \(A<\frac56\)

Vậy \(\frac{7}{12}<A<\frac56\)

Túi ko bt

\(D< \frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\)

\(=\frac{1}{1.2.3}-\frac{1}{98.99.100}=\frac{1}{6}-\frac{1}{98.99.10}< \frac{1}{6}\left(ĐPCM\right)\)

T.Anh 2K7(siêu quậy) làm đúng rồi. Làm nhanh và ngắn hơn tớ làm rất rất nhiều!!!

bai nay thi mk chiu that

\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)

\(\Rightarrow\)\(A< 1\) ( đpcm ) 

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

Sửa N=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)

Ta có : \(\frac{1}{2}< \frac{2}{3}\); \(\frac{3}{4}< \frac{4}{5}\); \(\frac{5}{6}< \frac{6}{7}\); ... ; \(\frac{99}{100}< \frac{100}{101}\)

\(\Rightarrow\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)hay M < N

b) M .N = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}=\frac{1.2.3.4.5.6...99.100}{2.3.4.5.6.7...100.101}=\frac{1}{101}\)

c) vì M < N nên M. M < M . N = \(\frac{1}{101}\)\(< \frac{1}{100}\)

\(\Rightarrow M< \frac{1}{10}\)

Đặt S bằng tổng dãy số trên.

=>S=3/4!+3/5!+.....+3/100!

=>S<3/4!+4/5!+.....+99/100!

=>S<1/3!-1/4!+1/4!-1/5!+.....+1/99!-1/100!

=>S<1/3!-1/100!

=>S<1/3!.Vậy S<1/3!
 

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