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please explain

 Apr 18, 2019
 #1
avatar+7725 
+1

\(\log_8\left(4^r\right)\\ =r \log_8\left(4\right)\\ =\dfrac{2}{3} r\)

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 Apr 18, 2019
 #2
avatar+7725 
+1

\(\displaystyle{ \sum_{r=k+1}^{60} (3r+2)\log_8\left(4^r\right)\\ =\sum_{r=k+1}^{60} \left(2r^2 + \dfrac{4}{3}r\right)\\ =2\sum_{r=k+1}^{60}r^2 + \dfrac{4}{3}\sum_{r=k+1}^{60} r\\ =2\left(73810 - \dfrac{k(k+1)(2k+1)}{6}\right)+\dfrac{4}{3}\left(1830-\dfrac{k(k+1)}{2}\right)\\ =150060-\dfrac{1}{3}\left(2k^3+3k^2+k\right)-\dfrac{2}{3}\left(k^2+k\right)\\ =150060-\dfrac{2}{3}k^3-\dfrac{5}{3}k^2-k\\ 150060-\dfrac{2}{3}k^3-\dfrac{5}{3}k^2-k > 106060\\ 2k^3+5k^2+3k-132000<0\\ k < 39.58 }\)

Greatest value of k is 39.

 Apr 18, 2019

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