Question

valuarea nr n=$\frac{ \sqrt{3+ \sqrt{12}+ \sqrt{27} + \sqrt{48}+...+ \sqrt{300} } }{ \sqrt{75} }$

Answer 1

$\it n = \dfrac { \sqrt{3} + \sqrt{12} + \sqrt{18} + .... + \sqrt{300}}{ \sqrt{75}} \rightarrow \\ \\ n = \dfrac{\sqrt{3} + 2 \sqrt{3} + 3 \sqrt{3} +... + 10 \sqrt{3}}{5 \sqrt {3}} \rightarrow \\ \\ n = \dfrac{ \sqrt{3}(1 + 2 +...+10)}{5 \sqrt{3}} \rightarrow \\ \\ n = \dfrac { \dfrac {10*(10+1)}{2}}{5} \rightarrow \\ \\ n = \dfrac{ 5*11}{5} \rightarrrow \\ \\ n = 11.$