Chứng minh $\frac{5}{4}$+ $\frac{5}{4^2}$ $\frac{5}{4^3}$ +..+$\frac{5}{4 mũ 2020}$ < $\frac{5}{3}$ 15/08/2021 Bởi aikhanh Chứng minh $\frac{5}{4}$+ $\frac{5}{4^2}$ $\frac{5}{4^3}$ +..+$\frac{5}{4 mũ 2020}$ < $\frac{5}{3}$
Ta có:5/ 4 + 5/ 4^2 + 5/ 4^3+ … + 5/ 4^2020 A= 5/ 4 + 5/ 4^2 + 5/ 4^3+ … + 5/ 4^2020 A/ 4= 5/ 4^2 + 5/ 4^3+ … + 5/ 4^2021 A/ 4= 5/ 4+ 5/ 4^2 + 5/ 4^3+ … + 5/ 4^2020 + 5/ 4^2021- 5/ 4 A/ 4= A + 5/ 4^2021- 5/ 4 3A/ 4= 5/ 4- 5/ 4^2021 A= 5/ 3- (5/ 4^2021 : 3/ 4) < 5/ 3 nha Bình luận
Giải thích các bước giải: \(C=\dfrac{5}{4}+\dfrac{5}{4^{2}}+\dfrac{5}{4^{3}}+…+\dfrac{5}{4^{2020}}\) \(\Leftrightarrow 4C=5+\dfrac{5}{4}+\dfrac{5}{4^{2}}+…+\dfrac{5}{4^{2019}}\) \(\Rightarrow 4C-C=(5+\dfrac{5}{4}+\dfrac{5}{4^{2}}+…+\dfrac{5}{4^{2019}})-(\dfrac{5}{4}+\dfrac{5}{4^{2}}+\dfrac{5}{4^{3}}+…+\dfrac{5}{4^{2020}})\) \(\Leftrightarrow 3C=5-\dfrac{5}{4^{2020}}\) \(\Leftrightarrow 3C<5\) \(\Leftrightarrow C<\dfrac{5}{3}\) Bình luận
Ta có:5/ 4 + 5/ 4^2 + 5/ 4^3+ … + 5/ 4^2020
A= 5/ 4 + 5/ 4^2 + 5/ 4^3+ … + 5/ 4^2020
A/ 4= 5/ 4^2 + 5/ 4^3+ … + 5/ 4^2021
A/ 4= 5/ 4+ 5/ 4^2 + 5/ 4^3+ … + 5/ 4^2020 + 5/ 4^2021- 5/ 4
A/ 4= A + 5/ 4^2021- 5/ 4
3A/ 4= 5/ 4- 5/ 4^2021
A= 5/ 3- (5/ 4^2021 : 3/ 4) < 5/ 3 nha
Giải thích các bước giải:
\(C=\dfrac{5}{4}+\dfrac{5}{4^{2}}+\dfrac{5}{4^{3}}+…+\dfrac{5}{4^{2020}}\)
\(\Leftrightarrow 4C=5+\dfrac{5}{4}+\dfrac{5}{4^{2}}+…+\dfrac{5}{4^{2019}}\)
\(\Rightarrow 4C-C=(5+\dfrac{5}{4}+\dfrac{5}{4^{2}}+…+\dfrac{5}{4^{2019}})-(\dfrac{5}{4}+\dfrac{5}{4^{2}}+\dfrac{5}{4^{3}}+…+\dfrac{5}{4^{2020}})\)
\(\Leftrightarrow 3C=5-\dfrac{5}{4^{2020}}\)
\(\Leftrightarrow 3C<5\)
\(\Leftrightarrow C<\dfrac{5}{3}\)