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Question

# If 2n∑r=0ar(x−2)r=2n∑r=0br(x−3)r and ak=1 for all k≥n, then bn is equal to

A
2(n+1)C2n+1
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B
2n+1Cn+1
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C
2n+1Cn+2
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D
None of these
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Solution

## The correct option is B 2n+1Cn+1Clearly, bn is the coefficient of (x−3)n in the expression 2n∑r=0br(x−3)r. ∴ bn=coefficient of (x−3)n in (2n∑r=0ar(x−2)r) ⋯(1) =coefficient of (x−3)n in n−1∑r=0ar(x−2)r+2n∑r=nar(x−2)r (∵ak=1 for all k≥n) =coefficient of (x−3)n in 2n∑r=n(x−2)r =coefficient of (x−3)n in [(x−2)n{(x−2)n+1−1(x−2)−1}] =coefficient of (x−3)n in ((x−2)2n+1−(x−2)nx−3) =coefficient of (x−3)n+1 in {(x−2)2n+1−(x−2)n} =coefficient of (x−3)n+1 in (x−2)2n+1 =coefficient of (x−3)n+1 in [(x−3)+1]2n+1 =coefficient of (x−3)n+1 in {2n+1∑r=0 2n+1Cr (x−3)r} = 2n+1Cn+1

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