Answers to exercises after class in the first volume of advanced mathematics

According to the first line, dn = (a+b) × d (n-1)-ab× d (n-2), so

dn-a×D(n- 1)= b×[D(n- 1)-a×D(n-2)]

D 1=a+b 0 = A+B, D2 = A 2+B 2+AB (where A 2 stands for the square of A).

So the sequence {dn-a× d (n- 1)} is a geometric series, the common ratio is b, and the first term is d2-a× d1= b 2.

Therefore, dn-a× d (n-1) = b 2× b (n-2) = b n.

Similarly, dn = (a+b) × d (n-1)-ab× d (n-2) gives dn-b× d (n-1) = a× [d (n-1)-b× d.

From dn-a× d (n- 1) = b n, dn-b× d (n-1) = a n.

dn=[a^(n+ 1)-b^(n+ 1)]/(a-b),n≥2

D 1 also satisfies the above formula, so dn = [a (n+1)-b (n+1)]/(a-b), n = 1, 2, ...