chứng minh 1/2-1/3+1/4-1/5+...-1/999+1/1000<2/5
Những câu hỏi liên quan
chứng minh: 499/1000<1/2^2+1/3^2+1/4^2+...+1/999^2<3/4
Chứng minh rằng \(\frac{1}{3\sqrt[2]{2}}+\frac{1}{4\sqrt[3]{3}}+\frac{1}{5\sqrt[3]{4}}+.....+\frac{1}{1000\sqrt[3]{999}}< \frac{11}{5}\)
a, tìm giá trị nguyên của n đê phân số : A = 3n+ 2 /n-1 được giá trị lớn nhất , giá trị nho nhất
b, chứng minh rằng : 1+ 1/3+1/5+...+1/999 -(1/2+1/4+...+1/1000)= (1/501+1/502+1/1000)
b) Vế trái = \(\left(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{1000}\right)\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{1000}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{999}+\frac{1}{1000}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{500}\right)\)
= \(\frac{1}{501}+\frac{1}{502}+...+\frac{1}{1000}\)= Vế phải
=> đpcm
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A=1/1×2+1/3×4+1/4×5+...1/999×1000
B=1/501×1000+1/502×999+...+1/999×502+1/1000×501
Tính A/B
tính B=(2016/1000+2016/999+2016/998+...+2016/501)/(-1/1*2+/-1/3*4+-1/5*6+...+-1/999*1000)
\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+...+\frac{2016}{501}}{\frac{-1}{1.2}+\frac{-1}{3.4}+...+\frac{-1}{999.1000}}=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)}\)
\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)
\(=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{500}\right)\right]}\)
\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+....+\frac{1}{999}+\frac{1}{1000}\right)}=\frac{2016}{-1}=-2016\)
Vậy B = - 2016
Bạn Xyz cho mik hỏi ở phần mẫu số tại sao lại có -2*(1/2+1/4+...+1/1000) vậy? Nó ở đâu ra thế?
Tính nhanh : \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt[1]{2}+\sqrt[2]{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt[3]{4}+\sqrt[4]{5}}+...+\frac{1}{\sqrt{999}+\sqrt{1000}}+\frac{1}{\sqrt[999]{1000}+\sqrt[1000]{1001}}\)
1/1*2*3+1/2*3*4+1/3*4*5+.........+1/998*999*1000
1. Chứng tỏ rằng:
a. 1/n + 1/n+1 = 1/n + 1/n+1
b. 1/1 . 1/2 +1/2 . 1/3+ 1/3 . 1/4+.......+ 1/998 . 1/999+ 1/999. 1/1000
a, Điều đương nhiên
b,\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{999.1000}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{999}-\frac{1}{1000}\)
= \(1-\frac{1}{1000}\)
= \(\frac{999}{1000}\)
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2-1+4-3+6-5+.............+1000-999
1/2*4+1/4*6+.............+1/98*1000
5/2*4+5/4*6+...........+5/98*100