cho D=1/3+2/32+...+101/1012
CMR D<3/4
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Ta có :
\(D=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+..............+\dfrac{100}{3^{100}}+\dfrac{101}{3^{101}}\)
\(3D=1+\dfrac{2}{3}+\dfrac{3}{3^2}+.............+\dfrac{100}{3^{99}}\)
\(3D-D=\left(1+\dfrac{2}{3}+\dfrac{3}{3^3}+.....+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+.......+\dfrac{101}{3^{101}}\right)\)
\(2D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6D=3+1+\dfrac{1}{3}+............+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6D-2D=\left(3+1+\dfrac{1}{3}+..........+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+......+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)\(4D=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{203}{3^{100}}< 3\)
\(\Rightarrow D< \dfrac{3}{4}\rightarrowđpcm\)
~ Học tốt ~
ta có C=1+4+4^2+........+4^100
4C=4+4^2+4^3+...+4^101
4C-C=3C=4^101-1
C=(4^101-1)/3
VẬY C<B/3
Cho M=1/2*2/3..............*99/100
N=2/3*3/4*...................*100/101
CMR : M<N
Tính: M*N
CMR;M<1/10
Bài a:
1.3.5......199 = 1.2.3.4......199.200/2.4.6.....200
= 1.2.3.4.........199.200/1.2.3.4....100.2100
=101.102.....200/2.2......2.2
=101/2 . 102/2 . 103/2 . ..... . 200/2
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}=A-\frac{101}{3^{101}}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(3A=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(\Rightarrow2A=3-\frac{1}{3^{100}}\Rightarrow A=\frac{3}{2}-\frac{1}{2.3^{100}}< \frac{3}{2}\)
\(\Rightarrow2D=A-\frac{101}{3^{101}}< A< \frac{3}{2}\Rightarrow D< \frac{3}{4}\)