YLL討論網

由 **bugzpodder** 於星期四八月 07, 2003 11:00 pm

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

由 **bugzpodder** 於星期四八月 07, 2003 10:44 pm

homogeneous recurrsive relations. i dont know it in chinese. convince yourself of identity:

ax^(n+2)+by^(n+2)=(x+y)(ax^(n+1)+by^(n+1))-xy(ax^n+by^n)

由 **Searchtruth** 於星期四八月 07, 2003 10:36 pm

完ㄌ..看不懂...>"<

由 **bugzpodder** 於星期四八月 07, 2003 10:32 pm

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

由 **Searchtruth** 於星期三八月 06, 2003 11:07 pm

... ...無言...
我今天放上來ㄉ題目就是今年台中一中校內數學競賽有獎徵答...此題是B卷~共有3卷~

由 **scsnake** 於星期三八月 06, 2003 10:56 pm

用Mathematica硬算...............20

由 **Searchtruth** 於星期三八月 06, 2003 7:20 pm

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?

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[quote="bugzpodder"]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[/quote]