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the pattern will be:
n=0, tn = 3
n=1, tn = 4
n=2, tn = 6
n=3, tn = 9
n=4, tn = 13
n=5, tn = 18
n=6, tn = 24
n=7, tn = 31
n=8, tn = 39
n=9, tn = 48
...

consider, n+1 divisible by 3 ...so, n = 2,5,8,11,14..... tn even/odd ..thus, A is not sufficient

consider, n-1 divisible by 4... n = 5,9,13,17,..... tn is even always even.
or, n = 1 + multiple of 4 (4,8,12....) ...is even always b/c multiple of 4 is always odd..
you might have noticed the pattern:
n=0 odd
n=1 even
n=2 even
n=3 odd
n=4 odd
n=5 even
n=6 even
n=7 odd
n=8 odd
........



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