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