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Question Number 101461 by Dwaipayan Shikari last updated on 02/Jul/20 Answered by MAB last updated on 02/Jul/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\frac{\sqrt{{n}}}{\:\sqrt{{r}}\left(\mathrm{3}\sqrt{{r}}+\mathrm{4}\sqrt{{n}}\right)^{\mathrm{2}} }=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\frac{{r}}{{n}}}\left(\mathrm{3}\sqrt{\frac{{r}}{{n}}}+\mathrm{4}\right)^{\mathrm{2}}…
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