Chứng minh\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{2003}{2004!}< 1\)
BÀI 1:TÍNH:
\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+....+\frac{100}{2^{100}}\)
BÀI 2: CHỨNG MINH RẰNG:
\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-.....-\frac{1}{2004^2}>\frac{1}{2004}\)
BÀI 3:THỰC HIỆN PHÉP TÍNH BẰNG CÁCH HỢP LÝ:
\(B=\frac{1}{3}+\frac{1}{6}.\left(1+2\right)+\frac{1}{9}.\left(1+2+3\right)+.....+\frac{1}{6045}.\left(1+2+3+....+2015\right)\)
Bài 4 :
a) Tính giá trị của biểu thức :
\(A=\left(\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right)\cdot\frac{31}{50}\)
b) Chứng tỏ rằng : \(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
Bài 1 :
1 . Tính :
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
2 . Biết : 13 + 23 + ... + 103 = 3025
Tính : S = 23 + 43 + 63 + .... + 203
Chứng minh rằng: \(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}
Chứng tỏ \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{2003}{2004!}<1\)
Bài 5 :
a) Tính giá trị của biểu thức :
\(A=\frac{\left(81,624:4\frac{4}{3}-4.505\right)^2+125\frac{3}{4}}{\left\{\left[\left(\frac{11}{25}\right)^2:0,88+3,53\right]^2-\left(2,75\right)^2\right\}:\frac{13}{25}}\)
b) Chứng minh rằng tổng :
\(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{4n-2}}-\frac{1}{2^n}+...+\frac{1}{2^{2002}-}-\frac{1}{2^{2004}}< 0,2\)
1/tính nhanh
a/1904.(2004-57)-2004.(1904-57)
b/\((\frac{1}{2}-1)\times(\frac{1}{3}-1)\times(\frac{1}{4}-1)\times...\times(\frac{1}{2004}-1)\)1)
c/\(\left(1+2^4+2^8\right):\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}+2^{11}\right)\)
A=\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2004}\right)\)
\(B=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}\cdot4\frac{1}{2}-2\cdot2\frac{1}{3}\right):\frac{7}{4}\)