log3(1+13)+log3(1+14)+log3(1+15)+⋯+log3(1+1242)=

log3(1+13)+log3(1+14)+log3(1+15)+⋯+log3(1+1242)=log3(3+13)+log3(4+14)+⋯+log3(242+1242)=log3(43)+log3(54)+⋯+log3243242=log343×54×65×⋯×242241×243242⇒log32433=log334=4log33=4×1=4

log3(1+13)+log3(1+14)+log3(1+15)+⋯+log3(1+1242)=

log3(1+13)+log3(1+14)+log3(1+15)+⋯+log3(1+1242)=log3(3+13)+log3(4+14)+⋯+log3(242+1242)=log3(43)+log3(54)+⋯+log3243242=log343×54×65×⋯×242241×243242⇒log32433=log334=4log33=4×1=4

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$\log_{3} (1 + \frac{1}{3}) + \log_{3} (1 + \frac{1}{4}) + \log_{3} (1 + \frac{1}{5}) + \dots + \log_{3} (1 + \frac{1}{242}) =$

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answer is B.

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Detailed Solution

$\begin{array}{l} \log_{3} (1 + \frac{1}{3}) + \log_{3} (1 + \frac{1}{4}) + \log_{3} (1 + \frac{1}{5}) + \dots + \log_{3} (1 + \frac{1}{242}) \\ = \log_{3} (\frac{3 + 1}{3}) + \log_{3} (\frac{4 + 1}{4}) + \dots + \log_{3} (\frac{242 + 1}{242}) \\ = \log_{3} (\frac{4}{3}) + \log_{3} (\frac{5}{4}) + \dots + \log_{3} \frac{243}{242} \\ = \log_{3} \frac{4}{3} \times \frac{5}{4} \times \frac{6}{5} \times \dots \times \frac{242}{241} \times \frac{243}{242} \\ \Rightarrow \log_{3} \frac{243}{3} = \log_{3} 3^{4} = 4 \log_{3} 3 = 4 \times 1 = 4 \end{array}$