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)

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<\(\frac{1}{2}\)

P = 1/4 + 1/16 + 1/36 + .. + 1/196 = 1/2² + 1/4² + 1/6² +...+ 1/12² + 1/14² 

xét tổng quát với số nguyên dương k ta có: 
(2k-1)(2k+1) = 4k² - 1 < 4k² = (2k)² => 1/(2k)² < 1/(2k-1)(2k+1) 
=> 2/(2k)² < 2 /(2k-1)(2k+1) = 1/(2k-1) - 1/(2k+1) (*) 

ad (*) cho k từ 1 đến 7 
2/2² < 1/1 - 1/3 
2/4² < 1/3 - 1/5 
... 
2/12² < 1/11 - 1/13 
2/14² < 1/13 - 1/15 
+ + cộng lại + + 
2/2² + 2/4² +...+ 2/14² < 1/1 - 1/15 < 1 
=> 2(1/2² + 1/4² +..+ 1/14²) < 1 => P < 1/2 (đpcm)

\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)

= \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

= \(\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)< \frac{1}{4}\left(1+1\right)=\frac{1}{2}\)

#ĐinhBa

kho lm giup minh voi

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 ~

Đặt \(A\)\(=\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có: \(A< \frac{1}{2^2-1}+\frac{1}{4^2-1}+...+\frac{1}{100^2-1}\)

 \(A< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\)

\(A< \frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(A< \frac{1}{2}.1\)( VÌ \(1-\frac{1}{101}< 1\))

\(A< \frac{1}{2}\)

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

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
Tick đúng nha bạn
 

1/4+1/16+1/36+...+1/196<1/4.2=1/2