Question

ta có B= \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+.....+\dfrac{2021}{4^{2021}}\)

so sánh B với \(\dfrac{1}{2}\)

giúp mik với mình cần gấp mai thi rồi

Answer 1

Lời giải:

\(B=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2021}{4^{2021}}\)

\(4B=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2021}{4^{2020}}\)

\(4B-B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)

\(3B=1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)

\(12B=4+1+\frac{1}{4}+...+\frac{1}{4^{2019}}-\frac{2021}{4^{2020}}\)

\(9B=4-\frac{6067}{4^{2021}}<4\Rightarrow B< \frac{4}{9}< \frac{1}{2}\)