Answer
$(\frac{3}{n})(\frac{1+1}{n})^{2}+(\frac{3}{n})(\frac{2+1}{n})^{2}+...+(\frac{3}{n})(\frac{n+1}{n})^{2}$=$(\frac{3}{n^{3}})\Sigma^{n}_{r=1} (r+1)^{2}$
Work Step by Step
$(\frac{3}{n})(\frac{1+1}{n})^{2}+(\frac{3}{n})(\frac{2+1}{n})^{2}+...+(\frac{3}{n})(\frac{n+1}{n})^{2}$=$(\frac{3}{n})(\frac{1}{n^{2}})(1+1)^{2}+(\frac{3}{n})(\frac{1}{n^{2}})(2+1)^{2}+...+(\frac{3}{n})(\frac{1}{n^{2}})(n+1)^{2}$
=$(\frac{3}{n})(\frac{1}{n^{2}})\Sigma^{n}_{r=1} (r+1)^{2}$
=$(\frac{3}{n^{3}})\Sigma^{n}_{r=1} (r+1)^{2}$
