A = 3/4 + 8/9 + 15/16 + ... + 399/400

A = 1 - 1/4 + 1 - 1/9 + 1 - 1/16 + ... + 1 - 1/400

A = (1 + 1 + 1 + ... +1) - (1/4 + 1/9 + 1/16 + ... + 1/400)

A = 19 - (1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20)

đặt b = 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20

có 1/2.2 < 1/1.2 ; 1/3.3 < 1/2.3 ; ... 1/20.20 < 1/19.20

=> b < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/19.20

=> b < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/19 - 1/20

=> b < 1 - 1/20

=> b < 1

mà A = 19 - b

=> A > 18

\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)

\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+.....+\frac{20^2-1}{20^2}\)

\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{20^2}\right)\)

\(>19-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{19\cdot20}\right)\)

\(=19-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{19}-\frac{1}{20}\right)\)

\(=19-\left(1-\frac{1}{20}\right)\)

\(>19-1=18\)

\(A=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+...+\frac{20^2-1}{20^2}\)

\(A=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+...+\frac{20^2}{20^2}-\frac{1}{20^2}\)

\(A=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\)

Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

\(B< 1-\frac{1}{20}< 1\)

\(B< 1\)

\(\Rightarrow A>19-1\)

\(\Rightarrow A>18\)

\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)

\(\Rightarrow M=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+....+\frac{20^2-1}{20^2}\)

\(\Rightarrow M=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+\frac{4^2}{4^2}-\frac{1}{4^2}+....+\frac{20^2}{20^2}-\frac{1}{20^2}\)

\(\Rightarrow M=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{20^2}\right)\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{20^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{19\cdot20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{19}{20}< 1\)

\(\Rightarrow A< 1\)

\(\Rightarrow M>18\)

A=(1-\(\frac{1}{4}\))+(1-\(\frac{1}{9}\))+(1-\(\frac{1}{16}\))+...+(1-\(\frac{1}{400}\)).

A=19-(\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\))

Ta thấy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}<1\)

=>A>19-1=18(đpcm)

\(a)\)\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)

\(M=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{400-1}{400}\)

\(M=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{400}\)

\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{400}\right)\)

\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)

Do từ 2 đến 20 có \(20-2+1=19\) nên : 

\(M=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\)

\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)

\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)

\(A>\frac{1}{2}-\frac{1}{21}\)

\(\Rightarrow\)\(M=19-A>19-\frac{1}{2}+\frac{1}{21}=18,5+\frac{1}{21}>8\)

\(\Rightarrow\)\(M>8\) ( đpcm ) 

Còn câu b) bn xem lại đề đi, nếu đề đúng thì mk sai :v 

Chúc bạn học tốt ~ 

\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}...+\frac{399}{400}\)

\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+\left(1-\frac{1}{25}\right)+...+\left(1-\frac{1}{400}\right)\)

\(=\left(1+1+1+....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{20^2}\right)\)

\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)

Đặt \(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{20^2}\)

\(< P=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{20\cdot21}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\)

\(=\frac{1}{2}-\frac{1}{21}\)

\(\Rightarrow M+N>19-\frac{1}{2}+\frac{1}{21}=\frac{37}{2}+\frac{1}{21}>8\)

b sai  đề.chừng nào chữa đề thì làm

Lời giải:

$A=(1-\frac{1}{4})+(1-\frac{1}{9})+(1-\frac{1}{16})+....+(1-\frac{1}{10000})$

$=(1+1+...+1)-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....+\frac{1}{10000})$

$=99-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....+\frac{1}{10000})< 99$

Đề thiếu. Bạn xem lại đề.

\(A\approx7.5\)

Tham khảo :
 

3​.98​.1615​.....100009999​

=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}=2.21.3​.3.32.4​.4.43.5​.....100.10099.101​

=\dfrac{\left(1.2.3.....99\right)}{\left(2.3.4.....100\right)}.\dfrac{\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)}=(2.3.4.....100)(1.2.3.....99)​.(2.3.4.....100)(3.4.5.....101)​

=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}=1001​.2101​=200101​
 

cộng mà bạn