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Question

 Numerically greatest term in the expansion of (2+3x)9, where x=32 is
  1. 3137
  2. 31272
  3. 31372
  4. 3127


Solution

The correct option is C 31372
We have, (2+3x)9=29(1+3x2)9

Now,
Tr+1Tr=10rr×3x2=9(10r)4r  [ x=32]
For Numerically greatest term,
Tr+1Tr19(10r)4r1r9013
So, Numerically greatest value occurs for  r=6 which is
T6+1=29×9C6(94)6=29×9C6(32)12=31372

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