Cmr :
1/3^2 - 1/3^4 +1/3^6 -...- 1/3^100< 0
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
Cho M=1/2*3/4*5/6*...*9999/10000 và N=2/3*4/5*6/7*...*10000/10001
a) CMR: M<N
b) CMR: M<1/100
S=1/5^2 - 2/5^3 + 3/5^4 -...+99/5^100-100/5^101
CMR S<1/3^6
bạn ấn vào đúng 0 sẽ ra kết quả, mình làm bài này rồi dễ lắm
A=1/2^2 + 1/4^2 + 1/6^3 + ... +1/100^2 . CMR A<1/2
\(A = \dfrac{1}{2^2} + \dfrac{1}{4^2} +\dfrac{1}{6^2} +...... +\dfrac{1}{100^2} \)
\(A = \dfrac{1}{1^2.2^2} +\dfrac{1}{2^2.2^2} +\dfrac{1}{2^2.3^2} + .......+\dfrac{1}{2^2.2^{50}}\)
\(A = \dfrac{1}{2^2}.(\) \( \dfrac{1}{1^2} + \dfrac{1}{2^2} +\dfrac{1}{3^2} +...... +\dfrac{1}{50^2}) \)
\(A < \dfrac{1}{2^2}.( \dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{49.50}\) \()\)
\(= \dfrac{1}{2^2}.(1-\dfrac{1}{2} + \dfrac{1}{2}-\dfrac{1}{3}+.......+\dfrac{1}{49}-\dfrac{1}{50})\)
\(= \dfrac{1}{2^2} . ( 1 - \dfrac{1}{50})\)
\(< \dfrac{1}{2^2} . 2 = \dfrac{1}{2}\)
CMR:(1+1/2+1/3+1/4+...+1/100)=1/2=2/3+3/4+...+99/100
CMR
100-(1/2+1/3+1/4+...+1/100)=2/3+3/4+...+99/100
100=10*10
100=1000:10
100 câu nói hay về cuộc sống
Bài 1 : Tính tổng
a) 1 *2 *3 + 2 * 3 *4 + 3 * 4 * 5 + ... + 2013 * 2014 * 2015 + 2014 * 2015 * 2016
b) 1 * + 3 * 4 + 5 * 6 + ... + 99 * 100
Bài 2 : CMR : 1^3 + 2^3 + 3^3 + ... + n^3 = ( 1 + 2 + 3 + ... + n )^2
Cho T=1/32 - 1/34 + 1/36 - ... - 1/3100. CMR T<0,1
\(T=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-...-\frac{1}{3^{100}}\)
\(\frac{1}{3^2}.T=\frac{1}{3^4}-\frac{1}{3^6}+\frac{1}{3^8}-...-\frac{1}{3^{102}}\)
\(\frac{1}{9}T+T=\frac{1}{3^2}-\frac{1}{3^{102}}\)
\(\frac{10}{9}T=\frac{1}{9}-\frac{1}{3^{102}}\Rightarrow T=\left(\frac{1}{9}-\frac{1}{3^{102}}\right).\frac{9}{10}=\frac{1}{10}-\frac{1}{3^{100}.10}< \frac{1}{10}=0,1\)
Vậy \(T< 0,1\)
cmr
100-(1+1/2+1/3+...+1/100)=1/2+2/3+3/4+....+99/100
\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)