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Please help me calculate the following sum

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

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3 Answers 3

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

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2  
This is the only answer so far that didn't outright solve the homework problem... –  Arkamis Sep 12 '12 at 18:46
    
@ Ha Hi: Complete answer. –  B. S. Sep 12 '12 at 18:47
    
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. –  Chris Leary Sep 12 '12 at 19:56

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

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

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:

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

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why (-1) downvoter? –  Iuli Sep 12 '12 at 18:47
3  
I didn’t do it, so I’m guessing, but probably because you solved a homework problem completely. –  Brian M. Scott Sep 12 '12 at 18:49
    
@BrianM.Scott Are you kidding ? I don't know one like this :) –  Iuli Sep 12 '12 at 18:50
    
Perhaps because your answer is a duplicate of the other two. –  Gigili Sep 12 '12 at 19:24
    
Really ? I don't care but I think is not ok how many vote. –  Iuli Sep 12 '12 at 19:31

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