Question

[doraemon]

cho B=1/4+1/4^2+...+1/4^2014 và C=1/52.(35/1.3+35/3.5+.....+35/103.105).so sánh B và C

Answer 1

[doraemon]

Giúp mk mn

Answer 2

[Xyz OLM]

Ta có B = \(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2014}}\)

=> 4B = \(1+\frac{1}{4}+...+\frac{1}{4^{2013}}\)

Lấy 4B trừ B theo vế ta có : 

4B - B = \(\left(1+\frac{1}{4}+...+\frac{1}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2014}}\right)\)

=> 3B = \(1-\frac{1}{4^{2014}}\)

=> B = \(\left(1-\frac{1}{4^{2014}}\right):3=\frac{1}{3}-\frac{1}{3.4^{2014}}\)

Lại có C = \(\frac{1}{52}\left(\frac{35}{1.3}+\frac{35}{3.5}+...+\frac{35}{103.105}\right)=\frac{1}{52}.\frac{35}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{103.105}\right)\)

\(=\frac{35}{104}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{103}-\frac{1}{105}\right)\)

\(=\frac{35}{104}.\left(1-\frac{1}{105}\right)=\frac{35}{104}.\frac{104}{105}=\frac{1}{3}\)

Vì \(\frac{1}{3}-\frac{1}{3.4^{104}}< \frac{1}{3}\Rightarrow B< C\)

Vậy B < C