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Questions  

In the expansion of x31x215, the constant term equals

a
15C9
b
0
c
- 15C9
d
15C11

detailed solution

Correct option is C

Tr+1, the (r+1)th term is the expansion of x3−1x215is Tr+1=15Crx315−r−1x2r=15Cr(−1)rx45−5rTo obtain constant term, we set 45−5r=0⇒r=9∴  Coefficient of constant term is  15C9(−1)9=−15C9.

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