If f(x)=1+x2−x3+...−x15+x16−x17, then the coefficient of x2 in f(x−1) is
A
826
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B
816
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C
822
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D
none of these
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Solution
The correct option is B816 The coefficient of x2 will be 2C2+3C2+4C2+5C2+...17C2 Applying nCr+nCr+1=n+1Cr+1 The above expression simplifies to [1+3C3+3C2+4C2+5C2+...17C2]−3C3 =[1+17C3+17C2]−1 =1+18C3−1 =18C3 =816