If the constant term in the expansion of ( [5]3x + 2x [3]5 )12, x 0, is 28 [5]3, then 25 is equal to :
- 724
- 742
- 693
- 639
Answer
(3) 693
Solution
Tr + 1 = 12Cr ( 31/5x )12 - r ( 2x51/3 )r
Constant term: power of x = r - (12 - r) = 0 r = 6
T7 = 12C6 36/5 2652 = 924 3 2625 31/5
= 924 34 25 28 [5]3 = 69325 28 [5]3
= 69325 25 = 693
Answer: (3)