Before diving into the statements, plug in a few numbers and see if there's a recurring pattern to the series:
T0 = 3
T1 = T0+1 = 3+1 = 4 = even
T2 = T1+2 = 4+2 = 6 = even
T3 = T2+3 = 6+3 = 9 = odd
T4 = T3+4 = 9+4 = 13 = odd
T5 = T4+5 = 13+5 = 18 = even
T6 = T5+6 = 18+6 = 24 = even
T7 = T6+7 = 24+7 = 31 = odd
T8 = T7+8 = 31+8 = 39 = odd
Notice a pattern? the series progresses in pairs of even and odd values: 2 Evens, 2 Odds, 2 Evens, 2 Odds.
The question stem asks you whether you can be sure that Tn is definitely even or definitely odd.
If Tn is always even for the n values allowed by a statement, the answer is "yes".
If Tn is always even for the n values allowed by a statement, the answer is "no".
The initial approach to such a Yes/No question will usually be one of "breaking the problem" - try to show that the statements allow more than one possible answer to the question - ony "yes", one "no":
(1) n+1 is divisible by 3. See what that means: n+1 could equal 3, 6, 9, leaving n with the values of 2, 5, 8. Now, our plug ins already show that if n=2, Tn is even, and the answer to the question stem is "yes", but if n=8, then Tn is odd and the answer to the question stem is "no". Therefore, Stat. (1)-->IS--> BCE
(2)n-1 is divisible by 4. This one is tricky to see, but its actually sufficient. n-1 could equal 4, 8, 12, leaving n itself as 5, 9, 13. Note that every multiple of 4 "closes" a loop of 2 even, 2 odd values of n:
T1 Even
T2 Even
T3 odd
T4 odd
("close loop")
T5 Even
T6 Even
T7 odd
T8 odd
("close loop".)
Etc.
Therefore, The next Tn will open a new "loop" with a new even value: T5, T9 and T13 and so on must be even values, as they are the next value after a multiple of 4. So the answer to the question is a definite Yes, and Stat. (2)-->S-->B.
