If n∑r=0Crxr , then the value of aC0−(a+d)C1+(a−2d)C2−(a−3d)C3+⋯+(−1)n(a−nd)Cn , equals
A
0
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B
(2a−n)2n(−1)n
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C
(−1)n(a+nd)2n
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D
None of these
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Solution
The correct option is A0 Simplifying, we get =a(nC0−nC1+nC2+...(−1)nnCn)−d(nC1−2nC2+3nC3+...(−1)ndnnCn) =[a(1−x)n+dd(1−x)ndx]|x=1 =0 Hence answer is 0.