Q.
The coefficient of x2 in the expansion of (1−x+2x2)(x+x1)10 is
4338
216
NTA AbhyasNTA Abhyas 2020Binomial Theorem
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Solution:
The required coefficient is the coefficient of x12 in (1−x+2x2)(x2+1)10 =(1−x+2x2)(1+10C1x2+10C2x4+……+10C9x18+x20)
Coefficient of x12 is 1×10C6+2×10C5=4×3×210×93×8×7+5×4×3×22×10××93×82×7×6 =210+504 =714