A=(1/2^2-1) (1/3^2-1) (1/4^2-1) .... (1/100^2-1)
A=(1/2^2-2^2/2^2) (1/3^2-3^2/3^2) ...... (1/100^2-100^2/100^2)
A=1-2^2/2^2 . 1-3^2/3^2 .... 1-100^2/100^2
A=-(2^2-1/2^2 . 3^2-1/3^2 ..... 100^2-1/100^2)
A=-(1.3/2^2 x 2.4/3^2 ..... 99.101/100^2)
A=-(1.3.2.4.....99.101/2.2.3.3.....100.100)
A=-[(1.2.3....99).(3.4.5.....101) / (2.3.4...100) . (2.3.4...100) ]
A=-101/200 < -1/2
Cho M = \(1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-....-\dfrac{1}{2^{10}}\) . So sánh M với \(\dfrac{1}{2^{11}}\)
\(\text{Cho A=}\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^4}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}};\)
Hãy so sánh A với \(\dfrac{1}{3}\)
So sánh:
a) A = \(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{49}}+\dfrac{1}{2^{50}}\) với 1
b) B = \(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\) với \(\dfrac{1}{2}\)
c) C = \(\dfrac{1}{4^1}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{999}}+\dfrac{1}{4^{1000}}\) với \(\dfrac{1}{3}\)
Cần gấp ạ ^^ Cảm ơn trước ^^
Cho \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\) và \(B=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{200^2}\). Khi đó \(\dfrac{A}{B}=...\)
so sánh :
A =\(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
và B = 2
cho A =\(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)
Chứng minh rằng A<\(\dfrac{1}{3}\)
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Cho B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{100^2}\). So sánh B với 3/4
a,\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+......+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\)
b,A=\(1+5+5^2+5^3+5^4+.....+5^{49}+5^{50}\)
c,A=\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)\)
d,A=\(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
Ai giúp mình thực hiện phép tính này với ạ?? Cảm ơn nhiều!!
