Let the coefficients of three consecutive terms Tr, Tr+1 and Tr+2 in the binomial expansion of (a + b)12 be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of ( [4]3 + [3]4)12. Then p + q is equal to:
- 295
- 283
- 299
- 287
Answer
(2) 283
Solution
Coefficients of Tr, Tr+1, Tr+2: 12Cr-1, 12Cr, 12Cr+1
G.P. 12Cr12Cr-1 = 12Cr+112Cr
13 - rr = 12 - rr + 1
(13 - r)(r + 1) = r(12 - r)
13 + 12r - r2 = 12r - r2 13 = 0, impossible
p = 0
General term of (31/4 + 41/3)12: Tk+1 = 12Ck 3 12 - k4 4 k3, 0 k 12
Rational 3 k and 4 (12 - k) 12 k k = 0, 12
q = 33 + 44 = 27 + 256 = 283
p + q = 0 + 283 = 283
Answer: (2)