TÍNH GIÁ TRỊ CỦA BIỂU THỨC:
1/4+1/16+1/36+1/64+1/100+1/121+1/144+...+1/10000 < 1/2
K
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BC
2
27 tháng 4 2018
\(Đ\text{ặt }S=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}\)
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(S=\frac{1}{2^2}\cdot\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1\cdot2};\text{ }\frac{1}{3^2}< \frac{1}{2\cdot3};\text{ }...;\text{ }\frac{1}{50^2}< \frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)
\(\Rightarrow\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}\cdot2\)
\(\Rightarrow S< \frac{1}{2}\) (ĐPCM)
SK
30 tháng 4 2018
Đặt \(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{100^2}\)
\(\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)
\(\Rightarrow4A< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow4A=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow4A< 2-\frac{1}{50}< 2\)
\(\Rightarrow4A< 2\Rightarrow A< \frac{2}{4}=\frac{1}{2}\)
=>a<1/2
TK
0
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)
MK
5
5 tháng 10 2017
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1x64}{2x64}+\frac{1x32}{4x32}+\frac{1x16}{8x16}+\frac{1x8}{16x8}+\frac{1x4}{32x4}+\frac{1x2}{64x2}+\frac{1}{128}\)
\(=\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\)
\(=\left(\frac{64}{128}+\frac{1}{128}\right)+\left(\frac{32}{128}+\frac{8}{128}\right)+\left(\frac{16}{128}+\frac{4}{128}\right)\)
\(=\frac{65}{128}+\frac{40}{128}+\frac{20}{128}\)
\(=125\)
TH
1
8 tháng 5 2019
A=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
=>A<\(\frac{1}{2.2}+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\)
=>A<\(\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\right)\)\(:2\)=\(\left(\frac{1}{2}-\frac{1}{14}\right):2\)<\(\frac{1}{2}\)
=>A<\(\frac{1}{2}\)
LT
7
TM
3 tháng 11 2017
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)\(-\frac{1}{64}\)
\(=1-\frac{32}{64}-\frac{16}{64}-\frac{8}{64}-\frac{4}{64}\)\(-\frac{2}{64}-\frac{1}{64}\)
\(=1-\left(\frac{32}{64}-\frac{16}{64}-\frac{8}{64}-\frac{4}{64}-\frac{2}{64}-\frac{1}{64}\right)\)
\(=1-\frac{1}{64}\)
\(=\frac{64}{64}-\frac{1}{64}\)
\(=\frac{63}{64}\)
DQ
0
NQ
4
25 tháng 4 2019
Bạn tham khảo nhé
A=14 +116 +136 +164 +1100 +1144 +1196 =122 +142 +162 +182 +1102 +1122 +1142
2A=222 +242 +262 +282 +2102 +2122 +2142
2A<12 +22.4 +24.6 +26.8 +28.10 +210.12 +212.14
2A<12 +12 −14 +14 −16 +16 −18 +18 −110 +110 −112 +112 −114
2A<12 +12 −114
2A<1−114
2A<1314
A<1328 <1428 =12 ( đpcm )
Vậy A<12
Chúc bạn học tốt ~
TB
3
9 tháng 8 2016
1/4 = 1/(2*2) < 1/(1*2) = 1/2 - 1/4
tương tự ta có
1/16 < 1/(2*4) = 1/4 - 1/8
1/36 < 1/(4*6) = 1/8 - 1/12
1/64 < 1/(6*8) = 1/12 - 1/16
1/100 < 1/(8*10) = 1/16 - 1/20
1/144 < 1/(10*12) = 1/20 - 1/24
1/196 < 1/(12* 14) = 1/24 - 1/28
cộng hết lại
=> 1/4 + 1/16 + ......+ 1/100 + 1/144 + 1/196 < 1/2 - 1/28 < 1/2 => đpcm
13 tháng 8 2016
ta có
1/4 = 1/(2*2) < 1/(1*2) = 1/2 - 1/4
tương tự ta có
1/16 < 1/(2*4) = 1/4 - 1/8
1/36 < 1/(4*6) = 1/8 - 1/12
1/64 < 1/(6*8) = 1/12 - 1/16
1/100 < 1/(8*10) = 1/16 - 1/20
1/144 < 1/(10*12) = 1/20 - 1/24
1/196 < 1/(12* 14) = 1/24 - 1/28
cộng hết lại
=> 1/4 + 1/16 + ......+ 1/100 + 1/144 + 1/196 < 1/2 - 1/28 < 1/2 => đpcm
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