Cmr : 1 + 1/1.2 + 1/1.2.3 + .....+ 1/1.2.3....n < 2
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Cmr : 1 + 1/1.2 + 1/1.2.3 + .....+ 1/1.2.3....n < 2
Cmr : 1 + 1/1.2 + 1/1.2.3 + .....+ 1/1.2.3....n < 2
cmr:\(\dfrac{1}{1.2}+\dfrac{2}{1.2.3}+....+\dfrac{2011}{1.2.3....2012}< 1\)
Lời giải:
\(A=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{2011}{1.2.3...2012}\)
\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{2012-1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...2011}-\frac{1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2...2012}< 1\)
Ta có đpcm.
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A= 1+1/1.2 + 1/1.2.3 +......+1/1.2.3.n <1+1/1.2 +1/1.3+......+1/k(k+1)
cmr:1/1.2+2/1.2.3+3/1.2.3.4+....+2011/1.2...2012<1
A=1+1/1.2+1+1/1.2.3+.......+1/1.2.3.n < 1+1/1.2+1/2.3+.......+1/k(k+1)
cmr:\(\dfrac{1}{1.2}+\dfrac{2}{1.2.3}+\dfrac{3}{1.2.3.4}+....+\dfrac{2011}{1.2...2012}< 1\)
CM:
1+\(\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 2\)
Đặt A = \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3....n}\)
Ta có: \(\frac{1}{1.2}=\frac{1}{1.2}\)
\(\frac{1}{1.2.3}=\frac{1}{2.3}\)
\(\frac{1}{1.2.3.4}< \frac{1}{3.4}\)
..............
\(\frac{1}{1.2.3....n}< \frac{1}{\left(n-1\right)n}\)
Cộng vế với vế ta được:
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1+1-\frac{1}{n}=2-\frac{1}{n}< 2\)(đpcm)
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Chứng minh bất đẳng thức sau:
\(1+\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+...+\dfrac{1}{1.2.3.....n}< 2\)