This is an awesome question!

First, notice that you can pull out a \(\frac{1}{3}\) from each of the terms in \(\frac{n+1}{3n}\).
For example:

\(a_1 = \frac{2}{3} = \frac{1}{3}*2\)

\(a_2 = \frac{3}{6} = \frac{1}{3}*\frac{3}{2}\)

\(a_3 = \frac{4}{9} = \frac{1}{3}*\frac{4}{3}\)
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The product of the first 53 terms would be:

\(a_1*a_2*a_3*.......a_{51}*a_{52}*a_{53}\)

In each term, we've pulled out a \(\frac{1}{3}\), and since we're multiplying them, this would result in:

\((\frac{1}{3})^{53}\)

Now notice what happened when we pulled out the \(\frac{1}{3}\) from the sequence in \(\frac{n+1}{3n}\).
Each term in the sequence turned into \(\frac{n+1}{n}\) (as seen above in the example).

So the product of our sequence looks like this:

\((\frac{1}{3})^{53}\) \( * (2*\frac{3}{2}*\frac{4}{3}*\frac{5}{4}.....\frac{53}{52}*\frac{54}{53})\)

All of these terms cancel except that last 54, so we're actually left with:

\((\frac{1}{3})^{53}\) \(* (54)\)

Breaking 54 down:

\((\frac{1}{3})^{53}\) \(* (3*3*3*2)\)

And this simplifies to:

\((\frac{1}{3})^{50}\) \(* (2)\)

Or:

\((\frac{2}{3^{50}})\)

Giving B as the answer

Given that \(a_n=\frac{n+1}{3n}\) and we need to find the product of the first 53 numbers in sequence S

\(a_1=\frac{1+1}{3*1}\) = \(\frac{2}{3*1}\)
\(a_2=\frac{2+1}{3*2}\) = \(\frac{3}{3*2}\)
\(a_3=\frac{3+1}{3*3}\) = \(\frac{4}{3*3}\)
\(a_4=\frac{4+1}{3*4}\) = \(\frac{5}{3*4}\)

So, Product of first 53 terms will be \(\frac{2}{3*1}* \frac{3}{3*2} * \frac{4}{3*3}*....*\frac{53}{3*52} * \frac{54}{3*53}\)
So, All numbers in numerator will cancel out apart from 54 and all numbers in denominator which are getting multiplied by 3 will cancel out

=> We will be left with \(\frac{54}{3*3*....*3}\) (3 in denominator will be 53 times one for each of the 53 terms)

= \(\frac{54}{3^53}\) = \(\frac{27*2}{3^53}\) = \(\frac{3^3*2}{3^53}\) = \(\frac{2}{3^50}\)

So, Answer will be B
Hope it helps!

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