Chứng minh A = \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....+\frac{1}{2011!}< 1\)\(1\)
Ai nhanh mình tick cho, giải rõ giùm mình nhé
\(y=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{2010}\right).\left(1+\frac{1}{2011}\right)\)
Tìm y
Giải nhớ có lời giải nhé
Ai làm nhanh mà đúng mình tick cho, hứa luôn
BẠN NÀO GIÚP MÌNH VỚI
CHỨNG MINH: \(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}< \frac{1}{2}\)
CÁC BẠN GIẢI ĐẦY ĐỦ HỘ MÌNH NHÉ, BẠN NÀO NHANH NHẤT MÌNH TICK CHO
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{1}{2^2}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\left(2-\frac{1}{7}\right)=\frac{1}{2}-\frac{1}{28}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\).
Chứng minh S=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+....................+\frac{1}{4008^2}+\frac{1}{4010^2}< \frac{1}{2}\)
Giúp mình nhé ai nhanh nhất mình tick
chứng minh rằng\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \)\(\frac{1}{2}\)
CÁC BẠN GIẢI RÕ GIÙM MK NHÉ, AI NHANH NHÂT VÀ CHÍNH XÁC NHẤT MK NHẤT ĐỊNH SẼ K CHO !!!
Ta có:\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)
\(=\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)\)\(< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)\(=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)
Vậy ............
Ta có: 1/3 + 1/31 + 1/35 + 1/37 + 1/47 + 1/53 + 1/61 < 1/3 + 3/31 + 3/47 < 1/3 + 3/30 + 3/45
= 1/3 + 1/10 + 1/15 = 1/3 + (1/30) * (3+2) = 1/3 + (1/0) * 5 = 1/3 + 1/6
= (1/6) * (2+1) = (1/6) * 3 = 1/2.
=> 1/3 + 1/31 + 1/35 + 1/37 + 1/47 + 1/53 + 1/61 < 1/2.
Ủng hộ mk nha mina^^
cac ban cho mk biet tai sao lai co phan so \(\frac{1}{30};\frac{1}{45}\)vay ???
Chứng minh
\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\)
Giúp mình nhé ai nhanh nhất mình tick
Nhận xét:
\(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}< \frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{1}{10}\)
\(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{45}+\frac{1}{45}+\frac{1}{45}=\frac{1}{15}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)
Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\) (Đpcm)
tính tổng:
\(S=1+\frac{1}{1+2}+\frac{1}{1+2+3}+..........+\frac{1}{1+2+3+.....+2011}\)
làm nhanh lên nhé mình cần gấp ! ai làm nhanh nhất mình tick cho 5 cái
\(\text{Công thức tổng quát: }\frac{1}{1+2+3+...+n}=\frac{2}{\left(n+1\right).n}\)
bạn thay vào òi làm tiếp ,phần tiếp theo dễ thui
Bài 1 ; So sánh
\(A=\frac{-1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)
\(B=\frac{1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
Mình cần gấp lắm ạ , Ai làm đúng và nhanh nhất mình tick cho
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
Chứng minh rằng:
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{2}\)
Giải nhanh nhanh giùm mình nha, 25/3 mình kiểm tra 45' rồi
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+.......+\frac{1}{100^2}<\frac{1}{2}\)
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+........+\frac{1}{100^2}\)<\(\frac{1}{0.2}+\frac{1}{2.4}+\frac{1}{4.6}+.......+\frac{1}{98.100}\)
\(S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<\frac{50}{100}=\frac{49}{100}<\frac{1}{2}\)
Vậy \(\frac{49}{100}<\frac{1}{2}\)
Ta có 1/22<1/2*3
1/42<1/3*4
. . .
1/1002<1/99*100
=> S<1/2*3+1/3*4+...+1/99*100
=> S<1/2-1/3+1/3-1/4+...+1/99-1/100
=>S<1/2-1/100
=>S<49/100
Mà 49/100<1/2
=>S<1/2
S = 1/2^2 + 1/4^2 + 1/6^2 + ... + 1/100^2
suy ra: 4*S = 1 + 1/2^2 + 1/3^2 + ... + 1/50^2
có: 1/2^2 = 1/2*2 < 1/1*2
1/3^2 = 1/3*3 < 1/2*3
1/50^2 = 1/50*50 <1/49*50
1+ 1/2^2 + 1/3^2 + ... + 1/50^2 < 1 + 1/1*2 + 1/2*3 + ... +1/49*50
4*S< 1 + 1 - 1/2 + 1/2 - 1/3 + ... + 1/49 - 1/50
4*S < 2 - 1/50 = 99/50
S < 99/50 : 4 = 99/50 * 1/4 = 99/200 < 100/200 = 1/2
vậy S < 1/2 (đpcm)
Chứng minh
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+........................+\frac{1}{78}+\frac{1}{79}+\frac{1}{80}< \frac{1}{2}\)
Giúp mình nhé ai nhanh nhất mình tick
ta có \(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}< \frac{1}{80}+\frac{1}{80}+..+\frac{1}{80}\)
ta có vế phải có 40 số , vế trái cũng có 40 số
VT=\(40\cdot\frac{1}{80}=\frac{40}{80}=\frac{1}{2}\)
do đó VT<1/2