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The sum of the coefficients of x2/3 and x-2/5 in the binomial expansion of (x2/3 + 12x-2/5 )9 is
(4) 21/4
Tr+1 = 9Cr (x2/3 )9 - r ( 12x-2/5 )r = 9Cr 12r x6 - 16r/15
since the power of x is 2(9 - r)3 - 2r5 = 6 - 16r15.
For x2/3: 6 - 16r15 = 23, so r = 5
For x-2/5: 6 - 16r15 = - 25, so r = 6
Sum = 9C5 125 + 9C6 126 = 12632 + 8464
= 6316 + 2116 = 8416 = 214
Answer: (4)
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