Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Chứng minh rằng :100- ( 1+1/2+1/3+...+1/100)=1/2+2/3+3/4+...+99/100
Chứng minh rằng:
1/3+1/3^2+1/3^3+...+1/3^99+1/3^100
chứng minh rằng :
1/2!.3! + 2/1!.2!.3! + ... + 99/98!.99!.100! < 1
Chứng minh rằng: 1/2!+2/3!+3/4!+...+99/100!<1
1.chứng minh rằng : \(\frac{1}{2}!+\frac{2}{3}!+\frac{3}{4}!+...+\frac{99}{100}!< 1\)
2. Chứng minh rằng :\(\frac{1.2-1}{2}+\frac{2.3-1}{3}+\frac{3.4-1}{4}+...+\frac{99.100-1}{100}< 2\)
CHỨNG MINH RẰNG
\(Q=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}< \frac{1}{2}\)
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
Chứng minh rằng:
a) A=1/3+1/(3^2)+1/(3^3)+...+1/(3^99)<1/2
b) B=3/(1^2*2^2)+5/(2^2*3^2)+7/(3^2*4^2)+...+19/(9^2*10^2)<1
c) C=1/3+2/(3^2)+3/(3^3)+4/(3^4)+...+100/(3^100)<3/4
Chứng minh rằng: 1/2! + 2/3! + ..... + 99/100! <1
