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Evaluate $\displaystyle{\sum_{i=1}^n \frac{i}{n^2}}$
$\displaystyle{\sum_{i=1}^n \frac{i}{n^2} = \frac{1}{n^2} \sum_{i=1}^n i = \frac{1}{n^2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{2n^2}}$
Evaluate $\displaystyle{\sum_{i=1}^n \left(\frac{1}{i} - \frac{1}{i+2}\right)}$
The sum telescopes. $\displaystyle{\sum_{i=1}^n \left(\frac{1}{i} - \frac{1}{i+2}\right) = \frac{1}{1} + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}}$
Evaluate $\displaystyle{\sum_{i=1}^n \left( \frac{6 \cdot 3^t}{4^t} \right)}$
Geometric series. $\displaystyle{\sum_{i=1}^n \left( \frac{6 \cdot 3^t}{4^t} \right) = \sum_{i=1}^n 6 \cdot \left( \frac{3}{4} \right)^t = \frac{6 \cdot 3 - 6 \cdot 3^{n+1}}{1-\frac{3}{4}} = 72(1-3^n)}$
Evaluate $\displaystyle{\sum_{i=1}^n \left( \frac{n}{n+i-1} - \frac{n}{n+i} \right)}$
The sum telescopes. $\displaystyle{\sum_{i=1}^n \left( \frac{n}{n+i-1} - \frac{n}{n+i} \right) = \frac{n}{n} -\frac{n}{2n} = 1 - \frac{1}{2} = \frac{1}{2}}$
Evaluate $\displaystyle{\sum_{i=1}^n \frac{3^t}{3 \cdot 2^{2t}}}$
Geometric series. $\displaystyle{\sum_{i=1}^n \frac{3^t}{3 \cdot 2^{2t}} = \sum_{i=1}^n \left( \frac{1}{3} \cdot \frac{3}{4} \right)^t = \frac{\frac{1}{4} - \frac{3^n}{4^{n+1}}}{1 - \frac{3}{4}} = 1 - \left(\frac{3}{4}\right)^n}$