Given a = Σ_(n=1) ^(24) (1/((√(n+1))+(√n))) then the value of a + (1/(log _a (bc)+1)) + (1/(log _b (ac)+1)) + (1/(log


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Question Number 103622 by bobhans last updated on 16/Jul/20

$G i v e n a = \underset{n = 1}{\sum^{24}} \frac{1}{\sqrt{n + 1} + \sqrt{n}} t h e n t h e v a l u e o f$ $a + \frac{1}{\log_{a} (b c) + 1} + \frac{1}{\log_{b} (a c) + 1} +$ $\frac{1}{\log_{c} (a b) + 1} = ?$
	Answered by OlafThorendsen last updated on 16/Jul/20

	$\frac{1}{\sqrt{n + 1} + \sqrt{n}} = \sqrt{n + 1} - \sqrt{n}$ $Then a = \underset{n = 1}{\sum^{24}} (\sqrt{n + 1} - \sqrt{n}) = \sqrt{25} = 5$ $5 + \frac{1}{\log_{5} (b c) + 1} + \frac{1}{\log_{b} (5 c) + 1} + \frac{1}{\log_{c} (5 b) + 1}$ $5 + \frac{1}{\frac{\ln (b c)}{\ln 5} + 1} + \frac{1}{\frac{\ln (5 c)}{\ln b} + 1} + \frac{1}{\frac{\ln (5 b)}{\ln c} + 1}$ $5 + \frac{\ln 5}{\ln (b c) + \ln 5} + \frac{\ln b}{\ln (5 c) + \ln b} + \frac{\ln c}{\ln (5 b) + \ln c}$ $5 + \frac{\ln 5}{\ln (5 b c)} + \frac{\ln b}{\ln (5 b c)} + \frac{\ln c}{\ln (5 b c)}$ $5 + \frac{\ln (5 b c)}{\ln (5 b c)} = 5 + 1 = 6$
	Commented by bobhans last updated on 16/Jul/20

	$c o o l l$
	Commented by bemath last updated on 16/Jul/20

	$w r o n g s i r \underset{n = 1}{\sum^{24}} \sqrt{n + 1} - \sqrt{n} = 4$
	Commented by OlafThorendsen last updated on 16/Jul/20

	$exact sir . you are right .$
	Answered by bemath last updated on 16/Jul/20

	$a + \frac{1}{\log_{a} (a b c)} + \frac{1}{\log_{b} (a b c)} + \frac{1}{\log_{c} (a b c)} =$ $a + \log_{a b c} (a b c) = a + 1$ $w h e r e a = \underset{n = 1}{\sum^{24}} \sqrt{n + 1} - \sqrt{n} = 5 - 1 = 4$ $t h e n \Leftrightarrow a + 1 = 4 + 1 = 5$
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