Chứng minh
A=1/3+2/3^2+3/3^3+4/3^4+...+101/3^101 < 3/4
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Chứng minh rằng: 1/3+2/3^2+3/3^3+4/3^4+..................+100/3^100+101/3^101<3/4
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{101}{3^{101}}\) (1)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{100}{3^{101}}+\frac{101}{3^{102}}\) (2)
Trừ (1) cho (2):
\(\frac{2}{3}A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{101}}-\frac{101}{3^{102}}=B-\frac{101}{3^{102}}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{101}}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{101}}+\frac{1}{3^{102}}\)
\(\Rightarrow\frac{1}{3}B+\frac{1}{3}-\frac{1}{3^{102}}=\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{101}}=B\)
\(\Rightarrow\frac{2}{3}B=\frac{1}{3}-\frac{1}{3^{102}}\Rightarrow B=\frac{1}{2}\left(1-\frac{1}{3^{101}}\right)=\frac{1}{2}-\frac{1}{2.3^{101}}\Rightarrow B< \frac{1}{2}\)
\(\Rightarrow A=\frac{3}{2}\left(B-\frac{101}{3^{102}}\right)< \frac{3}{2}B< \frac{3}{2}.\frac{1}{2}=\frac{3}{4}\)
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chứng minh : 1/3+2/3^2+3/3^3+..........101/3^101 < 3/4
S=1/3+2/3^2+...+101/3^101 chứng minh S<3/4
Chứng minh rằng 1/1*2+1/1*2*3+1/1*2*3*4+...+1/1*2*3*4*...*101 < 1
(1/2!+1/3!+1/4!+...+1/101!)
chứng minh rằng: 1/4+2/4^2+3/4^3+4/4^4+...+101/4^101 <4/9
Chứng minh:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{101}{3^{101}}
chứng minh rằng :
1/3+2/32+3/33+................+101/3101<3/4
1-1/2-2+2/3+3-3/4-4+4/5+5-5/6-...-100/101+101+101/102
S=1×2+2×3+3×4+4×5+...........+99×100
3S=1×2×3+2×3×(4-1)+3×4×(5-2)+4×5×(6-3)+............+99×100×(101-98)
3S=1×2×3+2×3×4-1×2×3+3×4×5-2×3×4+4×5×6-3×4×5+.............+99×100×101-98×99×100
3S=99×100×101
Tại sao 3S=99×100×101
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