A=2*(2+3+4+5+6+7+8+9+...........+99)+101+100
A = 2*4949+101+100
A=9898+101+100
A=10099
A=2*(2+3+4+5+6+7+8+9+...........+99)+101+100
A = 2*4949+101+100
A=9898+101+100
A=10099
a) 1+3-5-7+9+11-....-397-399
b) 1+2-3-4+5+6-....-99-100
c) 2-5+8-11+14-17+....+98-101
a) 1+3-5-7+9+11-......-397-399
b) 1+2-3-4+5+6-....-99+100
c) 2-5+8-11+14-17+....+98-101
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 = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
A = 1*2*3+3*4*5+5*6*7+...+99*99*100
Chứng tỏ rằng :1-1/2+1/3+1/4+...+1/99-1/200=1/101+1/102+...+1/199+1/200 .
Cho \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{101}{3^{101}}\),hãy so sánh A với \(\frac{3}{4}\)
1-2+3-4+5-6+...+99-100
Cho A=\(\dfrac{2}{1}.\dfrac{4}{3}.\dfrac{6}{5}.\dfrac{8}{7}.\dfrac{10}{9}...\dfrac{100}{99}\). Chứng minh rằng 12<A<13
Tính :
a) A=1*2+3*4+5*6+7*8+9*10+...+98*99
b) B=1*3+3*5+5*7+7*9+...+97*99
Giải câu nào cũng được (giúp mình đi)