$\Large \sum\limits_{x=1}^{2012} \frac{9^\frac{x}{2013}}{9^\frac{x}{2013}+3}=?$
I thought that if $\Large\frac{9^k}{9^k+3}$ was equal to $\Large\frac{9^{1-k}}{9^{1-k}+3}$ it would be easy to solve but apparently it isn't.
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Where did you find this? How do you know/why do you think there is a nice answer? – Bart Michels Jan 11 '14 at 16:22
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A friend asked me. He told me that the answer was 1005,25 – Zafer Cesur Jan 11 '14 at 16:39
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@ZaferCesur, do you mean the summation has two values? – lab bhattacharjee Jan 11 '14 at 16:56
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I meant 1005.25 but it was wrong anyway. Your solution is correct, thanks. – Zafer Cesur Jan 11 '14 at 16:59
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I get $1006$ as an answer? – Ali Caglayan Jan 11 '14 at 17:14
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Yes, it is correct. – Zafer Cesur Jan 11 '14 at 17:15
1 Answers
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HINT:
You are on the right way
If $\displaystyle f(k)=\frac{9^k}{9^k+3},$
$\displaystyle f(1-k)=\frac{9^{1-k}}{9^{1-k}+3}=\frac9{9+3\cdot9^k}=\frac3{9^k+3}$
$\displaystyle\implies f(k)+f(1-k)=1$
How many such pairs do we have ?
lab bhattacharjee
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