Pattern continued 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?

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