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 −1

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

a,\( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\)

= \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+...+ \dfrac{1}{196} < \dfrac{1}{2^2-1}+ \dfrac{1}{4^2-1}+ \dfrac{1}{6^2-1}+...+ \dfrac{1}{14^2-1}\)

= \( \dfrac{1}{1.3}+ \dfrac{1}{3.5}+ \dfrac{1}{5.7}+...+ \dfrac{1}{13.15}\)

= \( \dfrac{1}{2}(1- \dfrac{1}{3}+ \dfrac{1}{3}- \dfrac{1}{5}+ \dfrac{1}{5}- \dfrac{1}{7}+ \dfrac{1}{7}-...- \dfrac{1}{13}+ \dfrac{1}{13}- \dfrac{1}{15})\)

= \( \dfrac{1}{2}(1- \dfrac{1}{15})< \dfrac{1}{2}\)

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

b,A= \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)

\(=(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{40})+(\dfrac{1}{41}+...+1...\)
\(=(\dfrac{20}{20.21}+\dfrac{21}{21.22}+...+\dfrac{39}{39.40})+(40/...\)
\(20(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...\dfrac{1}{39.40})+40(\dfrac{1}{40}...\)
\(20(\dfrac{1}{20}-\dfrac{1}{40})+40(\dfrac{1}{40}-\dfrac{1}{60})>\dfrac{11}{15}\)
Lại có \(A<40(\dfrac{1}{20.21}+...\dfrac{1}{39.40})+60(\dfrac{1}{40.41}+...+...\)
\(=40(\dfrac{1}{20}-\dfrac{1}{40})+60(\dfrac{1}{40}-\dfrac{1}{60})<\dfrac{3}{2}\)

=> \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)

b)Tương tự câu a) nha bạn nhưng phải đổi là B=1/4+1/16+.....+1/196=1/2.2+1/4.4+.......+1/14.14

làm mấy bước tương tự câu a) cho đến khi ra B<1-\(\frac{1}{14}\)=\(\frac{13}{14}\)>\(\frac{7}{14}\)=\(\frac{1}{2}\)

        Bạn nến xem lại đề bài phần b) : B phải lớn hơn 1/2 chứ

khó hiểu lên thông cảm 

 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)