Find the coefficient of : a12 in the expansion of 4+2a−a2(1+a)3
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Solution
⟹(4−2a−a2)(1+a)−3 (1+a)−3:− Tr+1=n+r−1Cr(−1)r⋅ar For a12 Powers of ′a′ needed is :−12,11,10 T13=3+12−1C12(−1)12=14C12 T12=3+11−1C11(−1)11=−13C11 T11=12C10 ∴ Coefficient of a12 in above expression is :- ⟹4×14C12−2×13C11−12C10 ⟹142