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29 tháng 4 2018

\(N=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)

\(N=\dfrac{1}{2^1}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+\dfrac{1}{2^5}-\dfrac{1}{2^6}\)

\(2N=1-\dfrac{1}{2^1}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}-\dfrac{1}{2^5}\)

\(2N+N=1-\dfrac{1}{2^6}\)

\(N=\dfrac{1}{3}-\dfrac{1}{2^6.3}< \dfrac{1}{3}\left(đpcm\right)\)

21 tháng 7 2017

Ta có: \(VT=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(4VT=\dfrac{1}{2^2:2^2}+\dfrac{1}{4^2:2^2}+\dfrac{1}{6^2:2^2}+...+\dfrac{1}{100^2:2^2}\)

\(4VT=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

Lại có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(\Rightarrow4VT-1< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)(*)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}\) (**)

Từ (*) và (**) \(\Rightarrow4VT< 2-\dfrac{1}{50}\)

\(\Rightarrow VT< \dfrac{1}{2}-\dfrac{1}{200}< VP\Rightarrow\) đpcm

b) Ta có: \(2VT=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)

\(2VT+VT=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)

\(3VT=1-\dfrac{1}{64}< 1\)

\(\Rightarrow VT< \dfrac{1}{3}\) (đpcm)

22 tháng 7 2017

Thanks bạn nhìu nha!!!vuiyeu

8 tháng 5 2017

Sửa đề:

\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< 1\)

Ta có:

\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}\)

\(< \dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}< \dfrac{4}{4}< 1\)

25 tháng 5 2022

 

\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{128}-\dfrac{1}{256}\right)\)

\(A=1-\dfrac{1}{256}\)

\(A=\dfrac{255}{256}\)

 

 

25 tháng 5 2022

eo tự nhiên viết kh đc :v

17 tháng 12 2023

giải tri tiết nha

17 tháng 12 2023

1/5+45/9+1/2+1/3+1/2+1/9+1/15+1/99= ai trả lời đc đưa số tài khoản mik cho 100k

15 tháng 10 2021
Đáp án𝑠=15376
15 tháng 10 2021

Cả lời giải bn

BT1: CMR: a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\) b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\) c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\) d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\) e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\) f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\) BT2: Tính tổng a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\) b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\) BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\) CMR: 1 < S <...
Đọc tiếp

BT1: CMR:

a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)

b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)

c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\)

d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)

e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)

f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)

BT2: Tính tổng

a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)

b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\)

BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)

CMR: 1 < S < 2

1
22 tháng 3 2017

bài này có trong sách Nâng cao và Phát triển bạn nhé

10 tháng 8 2017

Giải:

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

\(\Leftrightarrow A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\)

\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}\)

Lấy vế trừ vế, ta được:

\(A-\dfrac{1}{2}A=\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^7}\)

\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^7}\)

\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}-\dfrac{1}{2^7}}{\dfrac{1}{2}}\)

\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}\left(1-\dfrac{1}{2^6}\right)}{\dfrac{1}{2}}\)

\(\Leftrightarrow A=1-\dfrac{1}{2^6}\)

Vậy \(A=1-\dfrac{1}{2^6}\).

Chúc bạn học tốt!!!

10 tháng 8 2017

Đặt:

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\)

\(2A=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\right)\)

\(2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)

\(2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\right)\)

\(A=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)

16 tháng 2 2022

\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

=>\(B=\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{6}{64}+\dfrac{2}{64}+\dfrac{1}{64}\)

=>\(B=\dfrac{32+16+6+2+1}{64}\)

=>\(B=\dfrac{63}{64}\)

16 tháng 2 2022

\(\dfrac{63}{64}\)