Given-a-n-1-24-1-n-1-n-then-the-value-of-a-1-log-a-bc-1-1-log-b-ac-1-1-log-c-ab-1-

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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