[(x32−x31+1)(x31+1)(x32−x31+1)−x(x−1)(x−1)(x+1)]10
=(x31+1−1−x211)10 =(x31−x2−1)10
tr+1=10Cr(x31)10−r(x2−1)r(−1)r
tr+1=10Cr(x310−65r)(−1)r
∵310−65r=−5
65r=650⇒r=10
Hence, 10C10(−1)10=1
The coefficient of x−5 in the binomial expansion of (x32−x31+1x+1−x−x21x−1)10 where x=0,1 is
Held on 9 Apr 2017 · Verified 6 Jul 2026.
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