1+2+3+...+100=1+3+2+.......+100
\(=\dfrac{\left(100+1\right)\left(100-1+1\right)}{2}=\dfrac{101\cdot100}{2}=5050\)
Chứng minh rằng :100- ( 1+1/2+1/3+...+1/100)=1/2+2/3+3/4+...+99/100
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 \(\dfrac{-1}{2}\)
tinh1/2(1+2)+1/3(1+2+3)+....+1/100(1+2+3+...+100)
tinh S= 1+1/2(1+2)+1/3(1+2+3)+...+1/100(1+2+3+...+100)
c/m
a/
1/2!+2/3!+3/4!+...+99/100!<1
b/
1*2-1/2!+2*3-1/3!+3*4-1/4!+...+99*100-1/100!<2
Tinh
A=1+1/2(1+2)+1/3(1+2+3)+...+1/100(1+2+3+...+100)
(1/100-12).(1/100-1/22).(1/100-1/32)....(1/100-1/202)
[1/100-1^2]ư.[1/100-(1/2)^2].[1/100-(1/3)^2]...[1/100-(1/20)^2]
Giải đầy đủ cho mik nha
1+ 1/2(1+2) + 1/3(1+2+3) +1/4(1+2+3+4) +....+1/100(1+2+3+..+100)
Ai giúp mk vs .....
Tính
A=1+1/2(1+2)+1/3(1+2+3)+...+1/100(1+2+3+...+100)
