If the last term in the binomial expansion of 21/3−12n is 135/3log3⁡8, the the 5th term from  the begging is   

If the last term in the binomial expansion of 21/312n is 135/3log38, the the 5th term from  the begging is  

 

  1. A

    210

  2. B

    420

  3. C

    105

  4. D

    None of these

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    Solution:

     Last term of  21/312n is 

    Tn+1=nCn21/3nn12n=nCn(1)n12n/2=(1)n2n/2

    Also, we have

    135/3log38=3(5/3)log323=25

     Thus, (1)n2n/2=25 (1)n2n/2=(1)1025 n2=5n=10

    Now T5=T4+1=10C421/3104124=10!4!6!21/36(1)421/24=210(2)2(1)22=210

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