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Giải thích các bước giải:
Đặt A= 1/4+1/16+1/36+1/64+1/100+1/144+1/196
= 1/2^2+ 1/4^2+ 1/6^2+….+ 1/16^2
= 1/2^2.( 1+ 1/2^2+ 1/3^2+…+ 1/8^2)
Ta có 1+ 1/2^2+ 1/3^2+…+ 1/8^2< 1+ 1/1.2+ 1/2.3+….7.8= 1+ 1-1/2+ 1/2- 1/3+….+ 1/7- 1/8
= 2- 1/8< 2
Nên ( 1+ 1/2^2+ 1/3^2+…+ 1/8^2)< 2
=> A< 1/2^2 nhân 2= 1/2
Vậy A< 1/2
Cho A = 1/2 .3/4.5/6.....199/200.Chứng tỏ rằng B mũ 2 <1/201.Bạn có làm dược ko ?
bài 2
| a;đặt biểu thức là S | |
| S < 1/1.2 + 1/2.3 + .......1/(n-1)n | |
| = 1- 1/2 +1 /2 -1/3+........ + 1/n-1 - 1/n | |
|
= 1 -1/n <1 |
|
| vậy S < 1 | |
\(A=\dfrac{1}{4}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{1010^2}\right)\)
1/2^2+1/3^2+...+1/2010^2<1/1*2+1/2*3+...+1/2009*2010=1-1/2010<1
=>A<1/4
\(N=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(N=\dfrac{1}{2^1}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+\dfrac{1}{2^5}-\dfrac{1}{2^6}\)
\(2N=1-\dfrac{1}{2^1}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}-\dfrac{1}{2^5}\)
\(2N+N=1-\dfrac{1}{2^6}\)
\(N=\dfrac{1}{3}-\dfrac{1}{2^6.3}< \dfrac{1}{3}\left(đpcm\right)\)
Đặt \(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)
Ta có:
\(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)
\(\Rightarrow A=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(\Rightarrow A< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)
\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)\)
\(\Rightarrow A< \dfrac{1}{4}.\dfrac{99}{50}\)
\(\Rightarrow A< \dfrac{99}{200}< \dfrac{1}{2}\)
Vậy \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}< \dfrac{1}{2}\) (Đpcm)
\(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(2-\dfrac{1}{50}\right)< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)
Sửa đề:
\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< 1\)
Ta có:
\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}\)
\(< \dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}< \dfrac{4}{4}< 1\)
Đặt \(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{196}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{13^2}\)
Đặt \(B=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{12\cdot13}\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{13^2}\)\(<\)\(B=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{12\cdot13}\left(1\right)\)
Mà \(B=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{12\cdot13}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{12}-\dfrac{1}{13}\)
\(=\dfrac{1}{2}-\dfrac{1}{13}< \dfrac{1}{2}\left(2\right)\). Từ \((1)\) và \((2)\) ta có:
\(A< B< \dfrac{1}{2}\Rightarrow A< \dfrac{1}{2}\) (Điều phải chứng minh)
Tuyệt quá bạn ơi