If ∑r=02n ar(x−2)r=∑r=02n br(x−3)r and ak = 1 for all k≥n, then bn is equal to

In the given equation, put x - 3 = y.∴ ∑r=02n ar(1+y)r=∑r=02n br(y)r⇒ a0+a1(1+y)+⋯+an−1(1+y)n−1+(1+y)n+(1+y)n+1+…+(1+y)2n=∑r=02n bryr                 [Using ak=1,∀k≥nEquating the coefficients of yn on both sides, we get nCn+n+1Cn+n+2Cn+…+2nCn=bn⇒  n+1Cn+1+n+1Cn+n+2Cn+…+2nCn=bn Using nCn=n+1Cn+1=1 ,  [ncr+ncr-1=(n+1)cr]⇒ bn=n+2Cn+1+n+2Cn+…+2nCnCombing the terms in similar way, we get⇒ bn=2nCn+1+2nCn=2n+1Cn+1