"If 1200∑n=139952005n+5200=ab, then find |a−b|" was answered by our community

sacateundisco8i3

Answered question

2022-02-27

\frac{1}{200} \sum_{n = 1}^{399} \frac{5^{200}}{5^{n} + 5^{200}} = \frac{a}{b}

, then find

| a - b |

Deichman82j

Beginner2022-02-28Added 2 answers

You have,

\frac{1}{200} (\sum_{n = 1}^{199} \frac{1}{5^{n - 200} + 1} + \frac{1}{2} + \sum_{n = 1}^{199} \frac{1}{5^{n} + 1})

= \frac{1}{200} (\sum_{n = 1}^{199} \frac{5^{200 - n}}{5^{200 - n} + 1} + \frac{1}{2} + \sum_{n = 1}^{199} \frac{1}{5^{n} + 1})

= \frac{1}{200} (\sum_{n = 1}^{199} \frac{5^{n}}{5^{n} + 1} + \frac{1}{2} + \sum_{n = 1}^{199} \frac{1}{5^{n} + 1})

= \frac{1}{200} (\sum_{n = 1}^{199} \frac{5^{n} + 1}{5^{n} + 1} + \frac{1}{2}) = \frac{399}{400}

Anderson Higgs

Beginner2022-03-01Added 5 answers

a very common trick among JEE summation questions is: pair up the first and last terms

\sum_{n = 1}^{399} \frac{5^{200}}{5^{n} + 5^{200}} = \frac{1}{2} + \sum_{n = 1, n \neq 200}^{399} \frac{5^{200}}{5^{n} + 5^{200}} = \frac{1}{2} + \sum_{n = 1}^{199} \frac{5^{200}}{5^{200} + 5^{n}} + \frac{5^{200}}{5^{200} + 5^{400 - n}}

and now an amazing thing happens

\frac{5^{200}}{5^{200} + 5^{n}} + \frac{5^{200}}{5^{200} + 5^{400 - n}} = 1

. So our sum is

\frac{1}{2} + 199 \cdot 1 = \frac{399}{2}

\Rightarrow \frac{a}{b} = \frac{1}{200} \cdot \frac{399}{2} = \frac{399}{400} \Rightarrow | a - b | = 1

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