Question

# The coefficient of x28 in the expansion of (1+x3−x6)30 is30C630C310

Solution

## The correct option is D 0(1+x3−x6)30=∑k1+k2+k3=3030!k1!k2!k3!(1)k1(x3)k2(−x6)k3=∑k1+k2+k3=3030!k1!k2!k3!(−1)k3x3k2+6k3 Now, for coefficient of x28 3k2+6k3=28⇒k2+2k3=283 As k2,k3∈W, so it is not possible Therefore, coefficient of x28 is 0.

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