cmr: 3/12x22+5/22x32+7/32x42+................+19/92x102<1
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Tính tổng S=(1/2x12 + 1/12x22 +1/22x32 +..........= 1/2002x2012)
Bài 1: CMR 3/1^2*2^2 + 5/2^2*3^2 + 7/3^2*4^2 + ....... + 19/9^2*10^2 bé hơn 1
Bài 2: CMR 1/3 + 2/3^2 Bài 1: CMR 3/1^2*2^2 + 5/2^2*3^2 + 7/3^2*4^2 + ....... + 19/9^2*10^2 bé hơn 3/4
Bài 3: Cho A= 1/1*2 + 1/3*4 + 1/5*6 + .... + 1/99*100. CMR 7/12 < A < 5/6
ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều
cmr : 3/1^2 . 2^2 + 5/2^2 . 3^2 + 7/3^2 . 4^2 + ....+ 19/9^2 . 10^2 < 1
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2-10^2}\)
\(=\) \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=\) \(1-\dfrac{1}{10^2}< 1\) ( đpcm )
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B1.Tìm A/B biết:
A=4/(7*31)+6/(7*41)+9/(10*41)+7/(10*57)
B=7/(19*31)+5/(19*43)+3/(23*43)+11/(23*57)
B2:CMR:
A=1/52+2/53+3/54+...+11/512 <1/16
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CMR : 3/1^2.2^2 + 5/2^2.3^2 + 7/3^2.4^2 + ... + 19/9^2.10^2 < 1
Ta có :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=1-\frac{1}{10^2}< 1\)
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cmr: 3/1^2.2^2 +5/2^2.3^2 +7/3^2.4^2+.....+19/9^2.10^2 <1
CHO A= 3/1^2.2^2+5/2^2.3^2+7/3^2.4^2+...+19/9^2.10^2CMR: A<1
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CMR:1×3×5×7×...×19=11/2×12/2×13/2×...×20/2
Ta có:
\(\dfrac{11}{2}.\dfrac{12}{2}.\dfrac{13}{2}.....\dfrac{20}{2}\\ =\dfrac{11.12.13.....20}{2^{10}}\\ =\dfrac{\left(11.12.13.....20\right)\left(1.2.3.....10\right)}{2^{10}\left(1.2.3.....10\right)}\\ =\dfrac{1.2.3.4.....20}{2.4.6.8.....20}\\ =\dfrac{\left(1.3.5.7.....19\right)\left(2.4.6.....20\right)}{\left(2.4.6.....20\right)}\\ =1.3.5.7.....19\)
=> Đpcm
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Cho H = 7/3 + 13/32 + 19/33 + . . . + 601/3100.
CMR: 3\(\frac{7}{9}\)< H <5
Cmr: \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
Ta có:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
= \(\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
= \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
= \(1-\dfrac{1}{10^2}\)
Mà \(1-\dfrac{1}{10^2}< 1\) nên:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) < 1 (đpcm).
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