n is an integer such that n ≥ 0. For n 0, the sequence tn is defined

Question

n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined as  t_n = t_{n-1} + n . If  t_0 = 3 , is  t_n  even?

(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4

ManojReddy · Accepted Answer

t_n = t_{n−1} +n
The above follows the below sequence
t0= Odd, t1=Even, t2=E,t3=O,t4=O,t5=E,t6=E,t7=O,t8=O,t9=E,t10=E. (E-Even, O-Odd)
So starting from t1 the sequence will be EEOOEEOOEEOOEEOO.......

(1) n + 1 is divisible by 3

n+1 being divisible by 3, n will be 2,5,8,11,14....
t2 = E, t5=E, t8=O
so we can't say whether tn is either odd or even.
 Insufficient

(2) n - 1 is divisible by 4
n-1 being divisible by 4, n will be 5,9,13,17,21....
t5=E,t9=E,t13=E,t17=E ....

Since the sequence is cyclic (EEOO). So tn will be even for all the values of n, where n-1 is divisible by 4.
 Sufficient

Ans: B

GMATinsight · Answer

tn = t_{n-1} + n

i.e.  t1 = t_0 + 1 = 3+1 = 4 
i.e.  t2 = t_1 + 2 = 4+2 = 6 
i.e.  t3 = t_2 + 3 = 6+3 = 9 
i.e.  t4 = t_3 + 4 = 9+4 = 13 
i.e.  t5 = t_4 + 5 = 13+5 = 18 
i.e.  t6 = t_5 + 6 = 18+6 = 24 
i.e.  t7 = t_6 + 7 = 24+7 = 31 
i.e.  t8 = t_7 + 8 = 31+8 = 39 
i.e.  t9 = t_8 + 9 = 39+9 = 48 
i.e.  t{10} = t_9 + 10 = 48+10 = 58 
i.e.  t{11} = t_{10} + 11 = 58+11 = 69 
and so on...

Eventually we can Observe that 2 terms are Even and the next two terms are odd and this cycle continues

Question : is  t_n  even?

Statement 1: n + 1 is divisible by 3

i.e. n+1 can be 3 or 6 or 9 or 12 etc
i.e. n can be 2 or 5 or 8 or 11 etc

t_2  is Even
 t_5  is Even
 t_8  is Odd
 t_{11}  is Odd.....i.e. INCONSISTENT answer

Hence,   NOT SUFFICIENT

Statement 2: n - 1 is divisible by 4

i.e. n-1 can be 4 or 8 or 12 etc
i.e. n can be 5 or 9 or 13 etc

The terms in this pattern will be at the difference of 4 and the property of all terms at difference of 4 terms is alike

t_5  is Even
 t_9  is Even
 t_{13}  is EVEN.....i.e. CONSISTENT answer

Hence,   SUFFICIENT

Answer: Option  B

bluesquare · Answer

Listing the first couple of iterations:
 n=0, t_0 = 3 
 n=1, t_1=1+1=2 
 n=2, t_2=2+2=4 
 n=3, t_3=4+3=7 
 n=4, t_4=7+4=11 
 n=5, t_5=11+5=16 
 n=6, t_6=16+6=22 
 n=7, t_7=22+7=29 
So you can see that when  n=even  the pattern is odd, even, odd, even, ..., or odd when n is a multiple of 4 and even otherwise, and when  n=odd  the pattern is even, odd, even, odd, ..., or odd when n+1 is a multiple of 4, and even otherwise.

1: if  n+1=3, n=2, t_2=4  but if  n+1=9,n=8,t_8=37  not sufficient
2: n-1 is divisible by 4 means that n+1 is not a multiple of 4, and so  t_n  will be even. sufficient. the answer is B.

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION: Sequence problems are often best approached by charting out the first several terms of the given sequence. In this case, we need to keep track of n, tn, and whether tn is even or odd. 2015-06-15_1618.png Notice that beginning with n = 1, a four-term repeating cycle of emerges for tn. Thus, a statement will be sufficient only if it tells us how n relates to a multiple of 4 (i.e. n = a multiple of 4 ± known constant). (1) INSUFFICIENT: This statement does not tell us how n relates to a multiple of 4. If n + 1 is a multiple of 3, then n + 1 could be 3, 6, 9, 12, 15, etc. This means that n could be 2, 5, 8, 11, 14, etc. From the chart, if n = 2 or n = 5, then tn is even. However, if n = 8 or n = 11, then tn is odd. (2) SUFFICIENT: This statement tells us exactly how n relates to a multiple of 4. If n – 1 is a multiple of 4, then n – 1 could be 4, 8, 12, 16, 20, etc. and n could be 5, 9, 13, 17, 21, etc. From the chart (and the continuation of the four-term pattern), tn must be even. The correct answer is B.

Divyadisha · Answer

t0= 3
t1= 3+1= 4 E
t2=4+2= 6 E
t3= 6+3=9 O
t4= 9+4= 13 O

We see that there is a sequence of EEOOEEOOEEOOEEOO

(1) n + 1 is divisible by 3
n can be 2, 5, 8 and hence it can be E or O
Not Sufficient.

(2) n - 1 is divisible by 4
n can be 5, 9, 13, 17 and it will always be E. 
Sufficient.

B is the answer

shmba · Answer

my approach:
Tn=T0+n(n+1)/2
this is probably the quickest approach to tackle the problem.

dan0015 · Answer

Hi. I did that too but can't understand how can we make sure that  /2 is odd or even.

Thanks!

MissionAdmit · Answer

t0=3
t1=3+1
t2=3+1+2
t3=3+1+2+3
t4=3+1+2+3+4..
tn=3+n(n+1)/2

3 is odd and thus n(n+1)/2 should be odd integer for tn to be even

I : N+1 = 3K , gives me nothing to confirm if (3k-1) 3k/2 is odd or even integer 
II : N-1=4J, (4J+1)(4J+2)/2 is (4J+1)(2J+1) - Odd X Odd which is always Odd. Thus sufficient

Answer is B

bumpbot · Answer

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n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined

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