A) \[\log 3\]
B) \[\log 2\]
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: C
Solution :
Key Idea In a given series there is a factorial in denominator, therefore it may be in the form of exponential series. \[\therefore \] \[1-\log 2+\frac{{{(\log 2)}^{2}}}{2!}-\frac{{{(\log 2)}^{3}}}{3!}+...\] \[={{e}^{-\log 2}}\] \[={{e}^{\log {{2}^{-1}}}}\] \[={{2}^{-1}}=\frac{1}{2}\]You need to login to perform this action.
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