What is the 30th term of this sequence (math GRE)?

Question

I am studying for the math subject GRE and came across the following problem from a previous exam (Form GR0568, found at "www.math.ucla.edu/~cmarshak/GRE1.pdf", question #25). The problem states:

25. Let $\{a_n\}_{n=1}^{\infty}$ be defined recursively by $a_1 = 1$ and $a_{n+1} = \left( \frac{n+2}{n} \right) a_n$ for $n \geq 1$. Then $a_{30}$ is equal to

(A) (15)(31)

(B) (30)(31)

(C) $\frac{31}{29}$

(D) $\frac{32}{30}$

(E) $\frac{32!}{30!2!}$

My approach was as follows: $a_2 = \frac{2+2}{2} = \frac{4}{2}$ therefore, we have

$$ a_{30} = \left( \frac{31}{29} \right) \left( \frac{30}{28} \right) ... \left( \frac{5}{3} \right) \left( \frac{4}{2} \right) \left( \frac{1}{1} \right) = \frac{31!}{3!29!} = \frac{(31)(30)}{6} = 31(5) $$

However, this is not even one of the answers! The answer is supposed to be (A), but I don't see how this is possible. If $a_1$ was not specially set to 1 but instead followed the same pattern to be $(1+2)/1 = 3$, then the extra factor of 3 would make the answer (15)(31) as expected, but unless there was a typo on a real subject math test that was used for years, I am going to go bald from all the head scratching I am doing for this problem.

Can anyone point out where I am going wrong on a seemingly simple problem?

Answer 1

A simple index mistake; $a_{\color{red}{n+1}}=\cdots$, so $a_2=\frac{3}{1}$ and not $\frac{4}{2}$. You should instead have $$a_{30}=\frac{31}{29}\frac{30}{28}\dots \frac{3}{1}=\frac{31!}{2!29!}=\frac{31\cdot 30}{2}=31\cdot 15 $$

Answer 2

Try to write the first few terms explicitly:

$a_1=1$

$a_2={3}/{1}$

$a_3={(3\times 4)}/{(1\times 2)}$

$a_4={(3\times 4\times 5)}/{(1\times 2\times 3)}$

$a_5={(3\times 4\times 5\times 6)}/{(1\times 2\times 3\times 4)}$

and note the telescoping that occurs between the numerator snd the denominator. It looks like you were doing this but kept the wrong product among the uncancelled terms in the denominator. So check again.