bugzpodder 寫到:this is recursive relation.
x,y are the roots of:
u^2-(x+y)u+xy=0
therefore
a1=3
a2=7
a3=16
a4=42
and a_n=(x+y)a_(n-1)-xya_(n-2)
16=7(x+y)-3xy
42=16(x+y)-7xy
solve for x+y and xy, then caculate a5 using formula
i made a sign error, i've fixed it
anyways, to solve it:
16*16=16*7(x+y)-48xy
42*7=16*7(x+y)-49xy
42*7-16*16=xy
xy=-38
x+y=-14
a5=ax^5+by^5=(x+y)a4-xya3=(-14)*42-(-38)*16=20