設a,b,x,y屬於R , ax+by=3 , ax2+by2=7
, ax3+by3=16 , ax4+by4=42
則ax5+by5之值為何?
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