Cho C= \(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+..+\frac{5}{4^{99}}\) Chứng minh C<5/3
Những câu hỏi liên quan
Cho C = \(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}.\)Chứng minh rằng C <\(\frac{5}{3}\)
C = \(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
= \(5\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
Đặt A = \(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\)
4A = \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\)
4A - A = \(\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
3A = \(1-\frac{1}{4^{99}}< 1\)
=> A < \(\frac{1}{3}\) (1)
Thay (1) vào C ta được:
\(C< 5\cdot\frac{1}{3}=\frac{5}{3}\)(đpcm)
Đúng 0
Bình luận (0)
Ta có:\(\frac{5}{4}\)< \(\frac{5}{3}\)Mà C = \(\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\)<\(\frac{5}{4}\)
\(\Rightarrow\)C < \(\frac{5}{3}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Cho C =\(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
Chứng minh C <\(\frac{5}{3}\)
Cho C = \(\frac{5}{4}\) + \(\frac{5}{4^2}\) + \(\frac{5}{4^3}\) + ... + \(\frac{5}{4^{99}}\)
Chứng minh rằng C < \(\frac{5}{3}\)
\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(4C=5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\)
\(4C-C=\left(5+\frac{5}{4}+...+\frac{5}{4^{98}}\right)-\left(\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\right)\)
\(3C=5-\frac{5}{4^{99}}\)
\(C=\frac{5-\frac{5}{4^{99}}}{3}\)
\(C=\frac{5}{3}-\frac{5}{4^{99}.3}< C\)
đpcm
Đúng 0
Bình luận (0)
Chứng minh rằng:a. frac{1}{3^2}+frac{2}{3^3}+frac{3}{3^4}+frac{4}{3^5}+...+frac{99}{3^{100}}+frac{100}{3^{101}} frac{1}{4}b.frac{1}{2}-frac{1}{4}+frac{1}{8}-frac{1}{16}+frac{1}{32}-frac{1}{64} frac{1}{3}c.frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{1}{16}d. frac{1}{5^2}-frac{2}{5^3}+frac{3}{5^4}-frac{4}{5^5}+...+frac{99}{5^{100}}-frac{100}{5^{101}} frac{1}{36}
Đọc tiếp
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
Cho S=\(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}\)
Chứng minh rằng \(S< \frac{1}{36}\)
Cho C=\(\frac{5}{4}\)+\(\frac{5}{4^2}\)+\(\frac{5}{4^3}\)+...+\(\frac{5}{4^{99}}\). Chứng minh:C<\(\frac{5}{3}\)
Chứng minh rằng \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\frac{5}{3^5}-.........+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
\(\frac{1}{2}-\frac{-2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\frac{4}{2^5}+...+\frac{99}{2^{99}}-\frac{100}{2^{100}}< \frac{2}{9}\)
Chứng minh
xem lại xem có sai đề bài không bạn ơi, sai thì sửa lại nhé
Đúng 0
Bình luận (0)
Chứng minh rằng : \(\frac{5}{6}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{11}{16}\)