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Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100
3A=1.2.(3-0) + 2.3.(4-1) +.....+99.100.(101-98)
3A=1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .....+99.100.101
3A=99.100.101
A=99.100.101/3=333300
Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100
3A= 3.(1.2 + 2.3 + 3.4 + ..... +99.100)
3A=1.2.(3-0) + 2.3.(4-1) +.....+99.100.(101-98)
3A=1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .....+99.100.101
3A=99.100.101
A=99.100.101/3=333300
đặt A = 1.2 + 3.4 + 4.5 +...+ 99.100
A=1.2+2.3+3.4+4.5+...+99.100
=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3
=1.2.3+2.3.﴾4‐1﴿+3.4.﴾5‐2﴿+4.5.﴾6‐3﴿+...+99.100.﴾101‐98﴿
=1.2.3+2.3.4‐1.2.3+3.4.5‐2.3.4+4.5.6‐3.4.5+...+99.100.101‐98.99.100
=1.2.3‐1.2.3+2.3.4‐2.3.4+3.4.5‐3.4.5+4.5.6‐4.5.6+...+99.100.101
=99.100.101=999900
=>A=999900:3=333300
Vậy A=333300
Ta có : \(\frac{1}{2.3}< \frac{1}{1.2}\)
\(\frac{1}{3.4}< \frac{1}{2.3}\)
\(\frac{1}{4.5}< \frac{1}{3.4}\)
...
\(\frac{1}{99.100}< \frac{1}{98.99}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\)
\(A< 1-\frac{1}{99}< 1\)
\(\Rightarrow A< 1\)
A \(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Vì \(\frac{49}{100}< 1\Rightarrow A< 1\)
Chúc bn hk tốt :>
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Sửa lại đề bài nha bạn:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
Chúc em học tốt nhé!
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
Gọi tổng trên là;A
A=9+99+999+........+999...9(20 số 9)
A=(10-1)+(100-1)+.......+(100...0(20 số 0)-1)
A=10+102+103+........+1020-(1+1+.........+1) 20 số 1
10A=102+103+.........+1021-200
10A-A=1021-10-200+20=1021-190
A=\(\frac{10^{21}-190}{9}\)
A = 1/2 - 1/3 + 1/3 -1/4 + 1/4 -1/5 + ...+ 1/98 - 1/99 + 1/99 - 1/100
A = 1/2 - 1/100 = 49/100