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Please help me calculate the following sum

$$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}$$

3 Answers 3

18

HINT: $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$

  • 2
    This is the only answer so far that didn't outright solve the homework problem...2012-09-12
  • 0
    @ Ha Hi: Complete answer.2012-09-12
  • 0
    For most homework problems, I think a hint is much more appropriate than a complete solution. Our job in these situations is not to do the students' work for them. But, I've been guilty of doing just that. So, I guess we just need to restrain ourselves a bit.2012-09-12
1

$$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}\\=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{99}-\frac{1}{100}\\=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}=0.49$$

1

$$\frac{1}{2 \cdot 3} =\frac{1}{2} - \frac{1}{3}$$ $$\frac{1}{3 \cdot 4}=\frac{1}{3} -\frac{1}{4}$$ $$\ldots$$ $$\frac{1}{99 \cdot 100}= \frac{1}{99}-\frac{1}{100}$$

So:

$$\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{99 \cdot 100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+ \ldots-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}.$$

general case:

$$\frac{1}{n \cdot (n+1)}=\frac{1}{n}-\frac{1}{n+1}.$$

  • 0
    why (-1) downvoter?2012-09-12
  • 3
    I didn’t do it, so I’m guessing, but probably because you solved a homework problem completely.2012-09-12
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    @BrianM.Scott Are you kidding ? I don't know one like this :)2012-09-12
  • 0
    Perhaps because your answer is a duplicate of the other two.2012-09-12
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    Really ? I don't care but I think is not ok how many vote.2012-09-12