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The sum of fraction is equal to

Solution: \(\begin{array}{l}\sum_{n=1}^{7}\frac{n(n+1)(2n+1))}{4}\end{array} \), \(\begin{array}{l}\frac{1}{4}\sum_{n=1}^{7}(2n^{3}+3n^{2}+n)\end{array} \), \(\begin{array}{l}\frac{1}{4}\left [2\sum_{n=1}^{7}n^{3}+3\sum_{n=1}^{7}n^{2}+\sum_{n=1}^{7}n \right ]\end{array} \), \(\begin{array}{l}\frac{1}{4}\left [2\times (\frac{7\times 8}{2})^{2}+3 \times (\frac{7\times 8\times 15}{2})+(\frac{7\times 8}{2}) \right... View Article

tan( + 2 ) is equal to

\(\begin{array}{l}\frac{\sqrt{2}sin\alpha }{\sqrt{1+cos2\alpha }} = \frac{1}{7}\end{array} \) and \(\begin{array}{l}\sqrt{\frac{1-cos2\beta }{2} }= \frac{1}{\sqrt{10}}\end{array} \), α, β ∈ (0, π/2 ), then tan(α... View Article