find the 4th term from the end in the expansion of (3/x2-x3/6)7

Dear student
Clearly, the given expansion contains 8 terms.So, 4th term from the end=(8-4+1)th=5th term from the beginning.Therefore,Required term=T5=T4+1=7C43x27-4-x364=7C33x23x364    7C4=7C3=7×6×53×2×133x6x1264=35x648
Regards

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