5 tháng 10 2018 lúc 22:22
Các câu hỏi tương tự
Cm
S= 1/22-1/24+1/26-...+1/22002-1/22004<0.2
CMR: 1:22 _ 1:24+1:26-...+1:24n-2-1:24n+...+1:22002-1:22004<0.2
CMR:S=1-1/2^2-1/3^2-...-1/2004^2>1/2004
CMR \(S=\frac{1}{2^2}-\frac{2}{2^4}+\frac{1}{2^6}-\frac{1}{2^8}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}< 0,2\)
1. C/m rằng
S = 1/2^2 - 1/2^4 + 1/2^6 - ... + 1/2^4n-2 - 1/2^4n + ... + 1/2^2002 - 1/2^2004 < 0,2
2. C/m rằng
B = 1 - 1/2^2 - 1/3^2 - 1/4^2 - ... - 1/2004^2 > 1/2004
Tính giá trị của các biểu thức sau 1) A1+2+2^2+...+2^{2015} 2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right) 3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89} 4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3} 5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}...
Đọc tiếp
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Chứng minh rằng tổng
S=1/2^2 -1/2^4 +1/2^6 - ...........+ 1/2^4n-2 -1/2^4n + ..........+ 1/2^2002 - 1/2^2004 nhỏ hơn 0,2
CMR:
S= \(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}< 0,2\)
[(1/2)+(1/3)+(1/4)+(1/5)+...+(1/2005)]/[(2004/1)+(2003/2)+(2002/3)+...+(1/2004)]