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The following pattern: $$\frac{3^{2/401}}{3^{2/401} +3}+\frac{3^{4/401 }}{3^{4/401} +3}+\frac{3^{6/401}}{3^{6/401} +3}+\frac{3^{8/401}}{3^{8/401} +3}$$

what will the result be if the pattern is continued $\;300\;$ times?

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    apologies for the poor text, I'm not sure what went wrong there it is meant to be 3 to the power of 2/4012012-12-09
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    Use base^{power} in LaTeX for exponentiation. 21^{76} is $21^{76}$.2012-12-09
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    ^ The {} are important...2012-12-09
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    thanks, im fairly new to LaTex2012-12-09
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    $n-$th term $$=\frac{3^{\frac{2,4,6,8\space n-th }{401}}}{3^{\frac{2,4,6,8\space n-th }{401}}+3}=\frac{3^{\frac{2n}{401}}}{3^{\frac{2n}{401}}+3}$$2012-12-09
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    I need the sum to the nth term2012-12-09
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    Your teacher seems to be a humorous man... :-)2012-12-09

2 Answers 2

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If you need the sum to the nth term, you're looking at computing the sum of the first 300 terms:

$$\sum_{k=1}^{300}\left(\large\frac{3^{\frac{2k}{401}}}{3^{\frac{2k}{401}}+3}\right)$$

To sum to the nth term, you need to compute:

$$\sum_{k=1}^{n}\left(\large\frac{3^{\frac{2k}{401}}}{3^{\frac{2k}{401}}+3}\right)= \sum_{k=1}^{n}\left(\large\frac{3\cdot 3^{\frac{2k-1}{401}}}{3\cdot\left(3^{\frac{2k-1}{401}}+1\right)}\right) = \sum_{k=1}^{n}\left(\large\frac{3^{\frac{2k-1}{401}}}{\left(3^{\frac{2k-1}{401}}+1\right)}\right)$$

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I highly doubt there is a simple closed form for this. WolframAlpha only returns (query) the decimal approximation: $130.6386445...$

Also, the inverse symbolic lookup calculator doesn't find anything...