GreenlightTestPrep wrote:
The sequence \(t_1\)\(, t_2, t_3, ..., t_n\) is such that \(t_n = \frac{n+1}{n+3}\) for all integers \(n≥1\). What is the product of the first \(40\) terms of the sequence?
A) \(\frac{1}{1806}\)
B) \(\frac{1}{1722}\)
C) \(\frac{1}{903}\)
D) \(\frac{1}{602}\)
E) \(\frac{1}{301}\)
As always, let's list a few terms from the beginning and end of the sequence to help us look for a pattern.
\(t_1 = \frac{1+1}{1+3}=\frac{2}{4}\)
\(t_2 = \frac{2+1}{2+3}=\frac{3}{5}\)
\(t_3 = \frac{3+1}{3+3}=\frac{4}{6}\)
.
.
.
\(t_{38} = \frac{38+1}{38+3}=\frac{39}{41}\)
\(t_{39} = \frac{39+1}{39+3}=\frac{40}{42}\)
\(t_{40} = \frac{40+1}{40+3}=\frac{41}{43}\)
ASIDE: I created this question to demonstrate the importance of NOT simplifying terms in a sequence. So, while we're listing terms in the sequence, we should avoid the temptation of immediately simplifying fractions like \(\frac{2}{4}\) and \(\frac{4}{6}\), because there are often cases in which it's much easier to see how to arrive at the correct answer by using the "raw" (non-simplified terms).So the product of the first 40 terms \(= (\frac{2}{4})(\frac{3}{5})(\frac{4}{6})...(\frac{39}{41})(\frac{40}{42})(\frac{41}{43})\)
\(= \frac{(2)(3)(4)(5).....(38)(39)(40)(41)}{(4)(5).....(38)(39)(40)(41)(42)(43)}\)
\(= \frac{(2)(3)}{(42)(43)}\)
[Notice that this simplification would have been much more difficult had we first simplified all of the terms in the sequence]\(= \frac{1}{(7)(43)}\)
\(= \frac{1}{301}\)
Answer: E
26 Sep 2021, 06:40 
71% (03:04) correct
