1/1.2+1/2.3+....+1/99.100
=1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100=100/100-1/100=99/100
Đặt A = 1/1.2 + 1/2.3 + ..... + 1/99.100
<=> A = 1/1 - 1/2 + 1/2 - 1/3 + .... + 1/99 - 1/100
<=> A = 1/1 - 1/100
<=> A = 1 - 1/100
=> A = 99/100
1/1.2+1/2.3+....+1/99.100
=1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100=100/100-1/100=99/100
Đặt A = 1/1.2 + 1/2.3 + ..... + 1/99.100
<=> A = 1/1 - 1/2 + 1/2 - 1/3 + .... + 1/99 - 1/100
<=> A = 1/1 - 1/100
<=> A = 1 - 1/100
=> A = 99/100
A=1/1x2 + 1/2x3 + ... + 1/99x100
Tính A
A = 1/1x2 + 1/2x3 +...+ 1/99x100
Tính A
A= 1/1x2 + 1/2x3 + 1/3x4 + .........+1/99x100
Tinh: 1/1x2+1/2x3+1/3x4+...+1/99x100
3x - (1/1x2+1/2x3+.....+1/99x100)=1/1x2x3+1/2x3x4+......+1/18x19x20
\(\frac{1}{1x2}+\frac{1}{2x3}+.....+\frac{1}{99x100}\)
Tính tổng dãy số có quy luật
S1=1+2+3+...+n
S2=1x2+2x3+....+99x100
chứng minh rằng:
1x2-1/2! + 2x3-1/3!+.....+99x100-1/100! <2
b, B = 1\(\frac{1}{1x2}+\frac{1}{2x3}+......+\frac{1}{99x100}\)
c, C = \(\frac{1}{1x2}+\frac{1}{2x3}+......+\frac{1}{n\left(n+1\right)}\)
d, D = 1 + 2 + 3 + ......+ n