Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
\(A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\)\(\frac{1}{3^{100}}\)
=> \(\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
=> \(A+\frac{1}{3}A=\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)\(+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
=>\(\frac{4}{3}A=\frac{1}{3}-\frac{1}{3^{101}}\)
=>\(A=\frac{1}{4}-\frac{1}{3^{100}.4}\)
tính nhanh (1+2+3+...+99+100).(1/2-1/3-1/7-1/9)(63.1,2-21.3,6)/1-2+3-4+...+99-100
A=(1+2+3+...+99+100)(1/2-1/3-1/7-1/9)(63.1,2-21.3,6)/1-2+3-4+...+99-100
S=1/1*2+1/2*3+1/3*4+...+1/99*100
S=1/1*3+1/3*5+1/5*7+....+1/99*101
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
(1/3+1/3^2+1/3^3+1/3^4).3^5+(1/3^5+1/3^6+1/3^7+1/3^8).3^9+.....+(1/3^97+1/3^98+1/3^99+1/3^100).3^101
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 \(\dfrac{-1}{2}\)
(1+2+3+4+5+....+99+100).(1/3-1/5-1/7-1/9).(0,3.12-21.6.3)
1/2+1/3+1/4+.......+1/100
Tinh:
A=1+5+5^2+...+5^120
B=3^100-3^99+3^98-3^97+...+3^2-3+1
C=1+7+7^2+...+7^2016
D=2^100-2^99+2^98-2^97+...+2^2-2
giup m voi
thanks
Tính nhanh:
A = \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
B = 13 + 23 + 33 + ... + 1253
