Hình như sai đề thì phải chứ mk làm ko đc !!!
A < 1/(1.2) + 1/(2.3) + 1/(3.4) + ...+ 1/(99.100)
<=> A< 1- 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + .. + 1/99 - 1/100
<=> A < 1 - 1/100 < 1 (đpcm)
So với thì đây
chung minh rang 1/3-2/3^2+3/3^3-4/3^4+....+99/3^99-100/3^100<3/16
chung minh rang A=\(\frac{1}{2}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+...+\frac{99}{2^{99}}-\frac{100}{2^{100}}<\frac{2}{9}\)
Chung minh rang: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
2, chung minh rang
a, \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}<\frac{1}{3}\)
b,\(\frac{1}{3}-\frac{2}{^{3^2}}+\frac{3}{3^4}+........+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
chumg minh rang 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16
chung minh rang 1-1/2^2-1/3^3-1/4^2-....-1/100^2>1/100
please,who can help me?
chung minh rang 1/3^2+1/4^2+1/5^2+...+1/100^2<1/2
chung minh rang
1/3 + 2/3^2 + ... + 100/3^100<3/4
chung minh rang 1\2 mu 2+1\3 mu 2+1\4 mu2+...+1\100 mu 2 < 1
