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Question Number 21643 by Joel577 last updated on 30/Sep/17

\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + \dots + \frac{2012}{1 + 2012^{2} + 2012^{4}}

Commented by Joel577 last updated on 30/Sep/17

\underset{n = 1}{\sum^{2012}} \frac{n}{1 + n^{2} + n^{4}}

= \underset{n = 1}{\sum^{2012}} (\frac{n}{(n^{2} - n + 1) (n^{2} + n + 1)})

= \underset{n = 1}{\sum^{2012}} (\frac{1 / 2}{n^{2} - n + 1} - \frac{1 / 2}{n^{2} + n + 1})

= (\frac{1 / 2}{1} - \frac{1 / 2}{3}) + (\frac{1 / 2}{3} - \frac{1 / 2}{7}) + (\frac{1 / 2}{7} - \frac{1 / 2}{13}) + \dots + (\frac{1 / 2}{2012^{2} - 2012 + 1} - \frac{1 / 2}{2012^{2} + 2012 + 1})

= \frac{1 / 2}{1} - \frac{1 / 2}{2012^{2} + 2013}

= \frac{1}{2} (1 - \frac{1}{2012^{2} + 2013})

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