Các câu hỏi tương tự
Chứng minh rằng
a, B = \(\frac{1\cdot2-1}{2!}+\frac{2\cdot3-1}{3!}+\frac{3\cdot4-1}{4!}+....+\frac{99\cdot100-1}{100!}< 2\)
c, C = \(\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{19}{9^2\cdot10^2}< 1\)
CM: \(\frac{1\cdot2-1}{2!}+\frac{2\cdot3-1}{3!}+\frac{3\cdot4-1}{4!}+...+\frac{99\cdot100-1}{100!}< 2\)
1 cmr
\(\frac{1\cdot2-1}{2!}\)+\(\frac{2\cdot3-1}{3!}\)+\(\frac{3\cdot4-1}{4!}\)+.............+\(\frac{99\cdot100-1}{100!}\)<2
HELP ME
Chứng tỏ \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}=\frac{4949}{19800}\)
tính:\(\frac{1\cdot98+2\cdot97+3\cdot96+...+97\cdot2+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+99\cdot100}\)
\(|x+\frac{1}{1\cdot2}|+|x+\frac{1}{2\cdot3}|+|x+\frac{1}{3\cdot4}|+...+|x+\frac{1}{99\cdot100}|=100x\)
Ai chơi Liên Quân mà có nick có trang phục Triệu Vân không chơi nữa cho mình được không.
Tính A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
Tính nhanh
B=\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
Tính :
\(A=\frac{1\cdot98+2\cdot97+3\cdot96+......+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+......+98\cdot99}\)
\(B=\frac{100-\left(1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+.........+\frac{99}{100}}\)