Sequence And Series Question 550
Question: If $ m,,n $ are the roots of the equation $ x^{2}-x-1=0 $ , then the value of $ \frac{( 1+m{\log_{e}}3+\frac{{{(m{\log_{e}}3)}^{2}}}{2,!,}+…\infty )( 1+n{\log_{e}}3+\frac{{{(n{\log_{e}}3)}^{2}}}{2,!,}+..\infty ),}{( 1+mn{\log_{e}}3+\frac{{{(mn{\log_{e}}3)}^{2}}}{2,!}+…..\infty )} $
Options:
A) 9
B) 3
C) 0
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
Numerator $ N={e^{m,{\log_{e}}3}}\times {e^{n{\log_{e}}3}} $ $ ={e^{{\log_{e}}3^{m}}}\times {e^{{\log_{e}}3^{n}}}=3^{m}\times 3^{n}={3^{m+n}} $ Denominator $ D={e^{mn{\log_{e}}3}}=3^{mn} $ whereas given $ m+n=1,,mn=-1 $
$ \therefore \frac{N}{D}=\frac{{3^{m+n}}}{3^{mn}}=\frac{3^{1}}{{3^{-1}}}=3^{2}=9 $ .
BETA
