Problem:
Given a finite sequence S=(a1,a2,…,an) of n real numbers, let A(S) be the sequence
(2a1+a2,2a2+a3,…,2an−1+an)
of n−1 real numbers. Define A1(S)=A(S) and, for each integer m,2≤m≤n− 1 , define Am(S)=A(Am−1(S)). Suppose x>0, and let S=(1,x,x2,…,x100). If A100(S)=(1/250), then what is x?
Answer Choices:
A. 1−22
B. 2−1
C. 21
D. 2−2
E. 22 Solution:
For every sequence S=(a1,a2,…,an) of at least three terms,
Thus for m=1 and 2 , the coefficients of the terms in the numerator of Am(S) are the binomial coefficients (0m),(1m),…,(mm), and the denominator is 2m. Because (rm)+(r+1m)=(r+1m+1) for all integers r≥0, the coefficients of the terms in the numerators of Am+1(S) are (0m+1),(1m+1),…,(m+1m+1) for 2≤m≤n−2. The definition implies that the denominator of each term in Am+1(S) is 2m+1. For the given sequence, the sole term in A100(S) is