Question

# If b=a+1, and n is a positive integer, find the value of bn−(n−1)abn−2+(n−2)(n−3)⌊2a2bn−4−(n−3)(n−4)(n−5)⌊3a3bn−6+....

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Solution

## By the Binomial Theorem, we see that1,n−1,(n−3)(n−2)⌊2,(n−5)(n−4)(n−3)⌊3,...are the coefficients of xn,xn−2,xn−4,xx−6,.... in the expansions of (1−x)−1,(1−x)−2,(1−x)−3,(1−x)−4,.... respectively. Hence the sum required is equal to the coefficient of xn in the expansion of the series11−bx−ax2(1−bx)2+a2x4(1−bx)3−a3x6(1−bx)4+...,and although the given expression consists only of a finite number of terms, this series may be considered to extend to infinity.But the sum of the series =11−bx÷(1+ax21−bx)=11−bx+ax2=11−(a+1)x+ax2, since b=a+1.Hence the given series =coefficient of xn in 1(1−x)(1−ax)=coefficient of xn in 1a−1(a1−ax−11−x)=an+1−1a−1.

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