1/2.3 + 1/3.4 + 1/4.5 + ... + 1/199.200
Tinh nhanh nhe
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Ta có: A=1.2+2.3+...+198.199+199.200
=>3A=1.2.3+2.3.3+...+198.199.3
+199.200.3
=>3A=1.2.3+2.3(4-1)+...+
198.199(200-197)+199.200(201-198)
=>3A=1.2.3+2.3.4-1.2.3+...+198.199.200
-197.198.199+199.200.201-198.199.200
=>3A=199.200.201
=>A=199.200.67
A=39800.67
A=2666600
3A =1.2.3 +2.3.(4-1) +3.4.(5-2) +4.5.(6-3)....+199.200.(201 -198)
= 1.2.3+2.3.4 -1.2.3 +3.4.5- 2.3.4 + 4.5.6 - 3.4.5 +......+ 199.200.201 -198.199.200
3A =199.200.201
A=199.200.67 =254600
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)
=\(1-\dfrac{1}{5}\)
=\(\dfrac{4}{5}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{200.201}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{200}-\frac{1}{201}\)
=\(\frac{1}{2}-\frac{1}{201}\)
=\(\frac{199}{402}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{200.201}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{200}-\frac{1}{201}\)
\(=\frac{1}{2}-\frac{1}{201}=\frac{199}{402}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..................+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(A=1-\frac{1}{200}\)
\(A=\frac{199}{200}\)
\(S=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.......+\frac{1}{49\cdot50}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.......+\frac{1}{49}+\frac{1}{50}\)
\(S=\frac{1}{2}-\frac{1}{50}\)
\(S=\frac{25}{50}-\frac{1}{50}\)
\(S=\frac{24}{50}=\frac{12}{25}\)
ai k mh mh k lại
k cho mh nha
S=1/2.3+1/3.4+1/4.5+....+1/49.50
=\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...........+\frac{1}{49x50}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{2}-\frac{1}{50}\)
=\(\frac{24}{50}\) mình cũng ko chắc đúng nhưng đây là cách giải của mình
A= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6
A= 1/2 - 1/3+ 1/3-1/4 + 1/4-1/5+ 1/5-1/6
A= 1/2- 1/6
A= 1/3
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{6}\)
\(\Rightarrow A=\frac{1}{3}\)