Question

Find the value of λ, where [.] denotes the greatest integer function, and

λ = $\left[\frac{10(\sqrt{27}+\sqrt{0}+\sqrt{27}+\sqrt{1}+\sqrt{27}+\sqrt{2}+\sqrt{27}+\sqrt{3}+...+\sqrt{27}+\sqrt{729})}{(\sqrt{27}+\sqrt{0}+\sqrt{27}-\sqrt{1}+\sqrt{27}-\sqrt{2}+\sqrt{27}-\sqrt{3}+...+\sqrt{27}-\sqrt{729})}\right]$

Answer 1

-19

Answer 2

• Let the expression be $\frac{N}{D}$.
• The numerator part is $N = 10 \sum_{k=0}^{729} (\sqrt{27} + \sqrt{k}) = 10 (730\sqrt{27} + \sum_{k=0}^{729} \sqrt{k})$.
• The denominator part is $D = (\sqrt{27}+\sqrt{0}) + \sum_{k=1}^{729} (\sqrt{27} - \sqrt{k}) = 730\sqrt{27} - \sum_{k=1}^{729} \sqrt{k}$.
• Let $A = 730\sqrt{27}$ and $Y = \sum_{k=1}^{729} \sqrt{k}$. Note $\sum_{k=0}^{729} \sqrt{k} = Y$.
• The expression becomes $\frac{10(A+Y)}{A-Y}$.
• We approximate $Y \approx \int_0^{729} \sqrt{x} dx = \frac{2}{3} (729^{3/2}) = 13122$.
• $A = 730\sqrt{27} = 2190\sqrt{3} \approx 3793$.
• Since $Y > A$, $A-Y$ is negative.
• The fraction is approximately $\frac{10(3793+13122)}{3793-13122} = \frac{10(16915)}{-9329} \approx -18.13$.
• Therefore, $\lambda = [-18.13] = -19$.