Carcass wrote:
Sequence S is the sequence of numbers \(a_1, a_2, a_3,\) ... , \(a_n\). For each positive integer n, the \(n\)th number \(a_n\) is defined by \(a_n=\frac{n+1}{3n}\) . What is the product of the first 53 numbers in sequence S?
A. \(\frac{2}{3^{53}}\)
B. \(\frac{2}{3^{50}}\)
C. \(\frac{2}{3^{49}}\)
D. \(\frac{3}{2^{50}}\)
E. \(\frac{2}{3^{25}}\)
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