The sequence t1, t2, t3, ..., tn, ... is such that tn = (n+1)/(n+3

Question

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}

Kinshook · Accepted Answer

Given: 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 . 
Asked: What is the product of the first  40  terms of the sequence?
 
The product of the first  40  terms of the sequence =  t_1*t_2*.....*t_{40} = \frac{2}{4}*\frac{3}{5}*\frac{4}{6}*...\frac{39}{41}*\frac{40}{42}*\frac{41}{43} = \frac{41!}{(43!/6)} = \frac{6}{43*42} = \frac{1}{301}

IMO E

Sajal2408 · Answer

Product = t1*t2*t3*t4*........*t39*t41 = (2/4)*(3/5)*(4/6)*......(39/41)*(40/42)*(41/43)

Denominator of each nth term will cancel of numerator of n+2 term.

So, Final result will have = (2*3)/(42*43) = 1/(7*43) = 1/301

Choice 'E'

100mitra · Answer

Correct Option E (1/301)
Product of 1st 40 term in sequence:
T(40) =

(n+1) =  2*3 *4*5........*39*40*41
(n+3) = 4*5*6......*39*40*41 *42*43

eliminate similar numbers 
Remaining numbers = (2*3)/(42*43) = 1/(7*43) = 1/301  Option E

Posted from my mobile device

BrentGMATPrepNow · Answer

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)}

= \frac{1}{(7)(43)}

= \frac{1}{301}

Answer: E

Hoozan · Answer

BrentGMATPrepNow  I fell for the trap of simplifying the question, BUT please could you tell me where I went wrong?

T1 = 1/2
T2 = 3/5
T3 = 2/3
T4 = 5/7
T5 = 3/4
T6 = 7/9
.
.
.
T40 = 41/43

So the product will be 1/2 * 3/5 * 2/3 * 5/7 * 3/4 * 7/9* . . . * 41/43

Now I tried seeing a pattern. Out here almost all the Denominators and Numerators get canceled: 1/  2   *  3 /  5   *   2  /  3   *   5  /  7   *   3  /4 *   7  /9* . . . * 41/43

So I thought that if we follow this pattern, apart from the first 3 in the Numerator and the last 43 in the Denominator, everything will get canceled out. Eventually, even the leftover 4, 9, and 41 as well

BrentGMATPrepNow · Answer

The problem with your approach is that is it's really hard to determine the pattern with regard to what gets cancelled and what doesn't. 
For example, your calculations one of the 3's and the numerator just left "uncanceled" BUT we can see that it cancels with a 9 in one of the denominators.

Hoozan · Answer

So whenever we get a sequence of terms we avoid simplifying the question?

BrentGMATPrepNow · Answer

I'm not suggesting that simplifying fractions will never be a good idea. 
However, for sequence questions, I suggest that we first leave fractions unsimplified to see if we spot a pattern. If we don't spot a pattern, THEN we should see what happens when we simplify the fractions.

Sabby · Answer

Hi,

In order to get an idea of the pattern, you can write down the 2 first terms and the last one :

1+1/1+3 * 2+1/2+3*...*40+1/40+3
2/4*3/5*...*41/43 
Therefore we get 2*3*4*5*...*41/4*5*6*7*...*43

This reminds me of permutations! Using this concept, it will be a fast and easy way to solve the question:
We get 41!/(43!/3!)

Now let's factorize : 
40!*41*3!/ 40!*41*42*43

Cancel out 40!*41 and you get 3!/42*43 or 6/42*43

Now Simplify and you get : 1/7*43 = 1/301

Answer E

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The sequence t1, t2, t3, ..., tn, ... is such that tn = (n+1)/(n+3

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The sequence t1, t2, t3, ..., tn, ... is such that tn = (n+1)/(n+3

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