MHB Hand Computing: Evaluate $\sum_{k=1}^{2013}f(k/2014)$

[magneto1]

Define:
\[
f(t) := \frac{7^t}{7^t + \sqrt{7}}.
\]
Without the aid of a computer or calculator, evaluate:
\[
\sum_{k=1}^{2013} f \left( \frac{k}{2014} \right).
\]
(Please show the work.)

 

[Saitama]

Define:
\[
f(t) := \frac{7^t}{7^t + \sqrt{7}}.
\]
Without the aid of a computer or calculator, evaluate:
\[
\sum_{k=1}^{2013} f \left( \frac{k}{2014} \right).
\]
(Please show the work.)

Spoiler

Notice that
$$f\left(\frac{k}{2014}\right)+f\left(\frac{2014-k}{2014}\right)=1$$
Hence, the sum is:
$$\sum_{k=1}^{2013} f \left( \frac{k}{2014} \right)=1006+f\left(\frac{1007}{2014}\right)=1006+\frac{\sqrt{7}}{2\sqrt{7}}=\boxed{1006.5}$$

 

[magneto1]

That's correct.

 

[lfdahl]

Spoiler

$$\sum_{k=1}^{2013}f\left ( \frac{k}{2014} \right )= \sum_{k=1}^{1007}\left \{ f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right ) \right \}=^* \sum_{k=1}^{1007}1=1007$$

\[(*). \;\;\; f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right )=\frac{7^{\frac{k}{2014}}}{7^{\frac{k}{2014}}+7^{\frac{1}{2}}}+\frac{7^{\frac{2014-k}{2014}}}{7^{\frac{2014-k}{2014}}+7^{\frac{1}{2}}}\\\\ =\frac{7^{\frac{k}{2014}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{1}{2} }\right )+7^{\frac{2014-k}{2014}}\left ( 7^{\frac{k}{2014}}+7^\frac{1}{2} \right )}{\left ( 7^{\frac{k}{2014}}+7^\frac{1}{2} \right )\left ( 7^{\frac{2014-k}{2014}}+7^\frac{1}{2} \right )} \\\\ =\frac{14 + 7^{\frac{1}{2}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{k}{2014}}\right )}{14 + 7^{\frac{1}{2}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{k}{2014}}\right )}=1\]

 

[magneto1]

Spoiler

$$\sum_{k=1}^{2013}f\left ( \frac{k}{2014} \right )= \sum_{k=1}^{1007}\left \{ f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right ) \right \}=^* \sum_{k=1}^{1007}1=1007$$

\[(*). \;\;\; f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right )=\frac{7^{\frac{k}{2014}}}{7^{\frac{k}{2014}}+7^{\frac{1}{2}}}+\frac{7^{\frac{2014-k}{2014}}}{7^{\frac{2014-k}{2014}}+7^{\frac{1}{2}}}\\\\ =\frac{7^{\frac{k}{2014}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{1}{2} }\right )+7^{\frac{2014-k}{2014}}\left ( 7^{\frac{k}{2014}}+7^\frac{1}{2} \right )}{\left ( 7^{\frac{k}{2014}}+7^\frac{1}{2} \right )\left ( 7^{\frac{2014-k}{2014}}+7^\frac{1}{2} \right )} \\\\ =\frac{14 + 7^{\frac{1}{2}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{k}{2014}}\right )}{14 + 7^{\frac{1}{2}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{k}{2014}}\right )}=1\]

Very close.

Spoiler

Consider the pair of $\sum_{k=1}^{1007}\left \{ f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right ) \right \}$ when $k=1007$.

 

[lfdahl]

Oh, my mistake!

Spoiler

\[\sum_{k=1}^{2013}f\left ( \frac{k}{2014} \right )= \sum_{k=1}^{1006}\left \{ f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right ) \right \}+f\left ( \frac{1007}{2014} \right )=^* \sum_{k=1}^{1006} \left \{ 1 \right \}+\frac{1}{2}=\frac{1}{2}2013\]