\sum^{n+1}_{i=1}3^{\frac{i}{n+1}} = 3^{\frac{1}{n+1}} + 3^{\frac{2}{n+1}} + ... + 3^{\frac{n+1}{n+1}} = 3^{\frac{1}{n+1}}\cdot ( 3^{\frac{1}{n+1} - \frac{1}{n+1}} + 3^{\frac{2}{n+1} - \frac{1}{n+1}} + ... + 3^{\frac{n+1}{n+1} - \frac{1}{n+1}}) = 3^{\frac{1}{n+1}}\cdot (1 + 3^{\frac{1}{n+1}} + ... + 3^{\frac{n}{n+1}}) = 3^{\frac{1}{n+1}}\cdot \frac{3^{\frac{n+1}{n+1}}-1}{3^{\frac{1}{n+1}}-1} = 3^{\frac{1}{n+1}}\cdot \frac{2}{3^{\frac{1}{n+1}}-1}} = \frac{2\sqrt[n+1]{3}}{\sqrt[n+1]{3}-1}