\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\cdot\cdot\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{19}{20}\)
\(=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot\cdot\cdot\cdot19}{2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot\cdot\cdot20}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot\cdot\cdot19\right)}{\left(2\cdot3\cdot4\cdot5\cdot\cdot\cdot19\right)\cdot20}\)
\(=\frac{1}{20}\)
a,B=1/2^2+1/3^2+...+1/8^2
suy ra B=1/2.2+1/3.3+1/4.4+....+1/8.8
mà 1/2.2<1/1.2;1/3.3<1/2.3;...;1/8.8<1/7.8
suy ra B<1/1.2+1/2.3+...+1/7.8
B<1-1/2+1/2-1/3+1/3-1/4+...+1/7-1/8
B<1-1/8<1 suy ra B <1
b,C=(1-1/2).(1-1/3)....(1-1/20)
C=1/2.2/3....19/20
C=1.2.3....18.19/2.3.4...19.20
C=1/20
(mình ko chắc vs hết quả phần b đâu nha)
a) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
mà : \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{8}\)
\(=1-\frac{1}{8}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1\)(đpcm)
