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Prove that $$\frac{1}{1⋅2}+\frac{1}{3⋅4}+\frac{1}{5⋅6}+....+\frac{1}{199⋅200}= \frac{1}{101}+\frac{1}{102}+\frac{1}{103}...+\frac{1}{200}$$

My Approach:

$T_{r}=\frac{1}{\left(2r\right)⋅\left(2r-1\right)}$

$\sum_{n=1}^{100}(T_{r}=\frac{1}{\left(2r\right)⋅\left(2r-1\right)}=\frac{\left(2r\right)-\left(2r-1\right)}{\left(2r\right)⋅\left(2r-1\right)}=\frac{1}{\left(2r-1\right)}-\frac{1}{2r})$

This gives:

$S=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.......+\frac{1}{197}-\frac{1}{198}+\frac{1}{199}-\frac{1}{200}$

which apparently leads to a dead end. Please help me out.

Mahmoud albahar
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Soumil Gupta
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    A question should be written in such a way that it can be understood even by someone who did not read the title. – José Carlos Santos May 06 '21 at 10:17
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    You are very close to the answer. Just add and subtract all the terms with even numbers in denominator. Group the terms so that you have sum of reciprocals from $n=1$ to $n=200$ minus twice the sum of even numbers in the reciprocals which cancels the first $100$ terms. – LM2357 May 06 '21 at 10:20
  • This should be helpful: https://math.stackexchange.com/q/2502982/668308 – one potato two potato May 06 '21 at 10:31

1 Answers1

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$S=1-\frac{1}{2}+\frac{1}{3}...-\frac{1}{200}=\sum_{n=1}^{200}\frac{1}{n}-2\sum_{n=1}^{100}\frac{1}{2n}=\sum_{n=101}^{200}\frac{1}{n}$

one potato two potato
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Yamato
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