Passage If n is the positive integer and if (1+4x+4x2)n=2n∑r=0arxr, where ar are real numbers. On the basis of above information answer the following questions. The value of 2n∑r=1a2r−1 is
A
9n−1
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B
9n+1
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C
9n−2
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D
9n+2
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Solution
The correct option is A9n−1 (1+4x+2x2)n (1+2x)2n ...(i) =∑2n0arxr Substituting x=1 in equation (i), we get 32n ...(a) Substituting x=−1 in equation (i), we get (−1)2n ...(b) Subtracting b from a, we get 2∑n0a2r−1=32n−1 =9n+1