-2+3-4+5-6+...+99-100
= (1+3+5+...+99) - (2+4+6+...+100)
tu 1 den 99 co : (99-1):2+1=50
1+..+99 = (1+99)x 50:2= 2500
tu 2 den 100 co : (100-2);2+1=50
2+...+100 = (100 +2) x 50:2=2550
=> A= 2500-2550=-50
A = 1*2*3+3*4*5+5*6*7+...+99*99*100
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 = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
ai giải giùm với ạ
P = 91+2+3+...+100)(1/2-1/3-1/7-1/9)(63 x 1.2 - 21 x 3.6) tất cả trên 1-2+3-4+5-6+...+99-100
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{10}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(\dfrac{\left(1+2+3+...+99+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}\)
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
Rút gọn ;
D = \(\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-\dfrac{1}{5^4}+\dfrac{1}{5^5}-\dfrac{1}{5^6}+...+\dfrac{1}{5^{99}}-\dfrac{1}{5^{100}}+\dfrac{1}{6.5^{100}}\)
CMR : 1/3 - 2/3^2 + 3^3 - 4/3^4 + .... + 99/3^99 - 100/3^100 < 3/16
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
