In the expansion of ( [3]2 + 1 [3]3 )n, n N, if the ratio of 15th term from the beginning to the 15th term from the end is 16, then the value of nC3 is
- 4960
- 2300
- 1040
- 4060
Answer
(2) 2300
Solution
Let a = 21/3, b = 3-1/3.
15th term from the beginning: T15 = nC14 an - 14 b14
15th term from the end = 15th term of (b + a)n = nC14 bn - 14 a14
Ratio = an - 28bn - 28 = ( ab )n - 28 = (21/3 31/3 )n - 28 = 6 n - 283
6 n - 283 = 16 = 6-1
n - 28 = -3 n = 25
25C3 = 25 24 236 = 2300
Answer: (2)