If f(x)=1−x+x2−x3+⋯−x15+x16−x17, then the coefficient of x2 in f(x−1) is
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Solution
f(x)=1−x+x2−x3+⋯−x15+x16−x17 ⇒f(x)=1−x181+x ⇒f(x−1)=1−(x−1)181+(x−1) Therefore, required coefficient of x2 is equal to the coefficient of x3 in 1−(1−x)18, which is given by 18C3=816