Question

在f(m,n)中,m、n、f(m,n)均為非負(fù)整數(shù),且對(duì)任意的m、n有f(0,n)=n+1,f(m+1,0)=f(m,1),f(m+1,n+1)=f(m,f(m+1,n)),

(1)求f(1,0)的值;

(2)求f(2,n)關(guān)于n的表達(dá)式;

(3)求f(3,n)關(guān)于n的表達(dá)式.

Answer 1

(1)∵f(0,n)=n+1,

令n=1,則f(0,1)=2.

又f(m+1,0)=f(m,1),令m=0,則f(1,0)=f(0,1)=2.

(2)∵f(m+1,n+1)=f(m,f(m+1,n)),

∴f(1,n)=f(0,f(1,n-1))=f(1,n-1)+1.

∴{f(1,n)}是以f(1,0)=2為首項(xiàng),公差d=1的等差數(shù)列,

且f(1,n)為第n+1項(xiàng).

∴f(1,n)=n+2.

又f(2,n)=f(1,f(2,n-1))=f(2,n-1)+2,

∴{f(2,n)}是以f(2,0)為首項(xiàng),公差d=2的等差數(shù)列.

∴f(2,n)=f(2,0)+(n+1-1)d=2n+f(2,0).

又由f(m+1,0)=f(m,1) f(2,0)=f(1,1)=3,

∴f(2,n)=2n+3.

(3)∵f(3,n)=f(2,f(3,n-1))=2·f(3,n-1)+3,

∴f(3,n)+3=2(f(3,n-1)+3).

∴{f(3,n)}是以f(3,0)+3= f (2,1)+3=8為首項(xiàng),公比為2的等比數(shù)列.

∴f (3,n)=8·2ⁿ-3=2ⁿ⁺³-3.