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

  • 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
    Really ? I don't care but I think is not ok how many vote.2012-09-12