A special type of partial gate number series problem Finding the Master 30

You may wish to set the original cup of water with a concentration of , and the original cup of wine as , where n is the nth time to pour into each measuring cup.

a(0)=0%,b(0)=100%。

b(n+1)=(3a(n)+10b(n))/13=3/13a(n)+10/13b(n);.1)

a(n+1)=(7a(n)+3b(n+1))/10=7/10a(n)+3/10b(n+1)=7/10a(n)+3/10(3/13a(n)+10/13b(n))

10/13a(n)+3/13b(n)..2)

From 2) obtains: b(n)=(13a(n+1)-10a(n)) 3, then b(n+1)=(13a(n+2)-10a(n+1)) 3, substitute 1) formula, obtain: (13a(n+2)-10a(n+1)) 3=3 13a(n)+10 13*(13a(n+1)-10a(n)) 3, simplify to obtain:

13a(n+2)-20a(n+1))+7a(n)=0；

13r^2-20r+7=0,r1=1,r2=7/13;

a(n)=c1*1^n+c2*(7/13)^n=c1+c2*(7/13)^n

Because: a(0)=0,a(1)=10 13a(0)+3 13b(0)=3 13.

So: c1+c2=0, c1+7 13*c2=3 13, get: c1=1 2, c2=-1 2

So a(n)=1 2-1 2*(7 13) n,b(n)=(13a(n+1)-10a(n)) 3=1 2+1 2*(7 13) n(n=0,1,2,3,..

So a(n)<1 2,b(n)>1 2.

When n->, a(n)=b(n)=1 2;

Therefore, no matter how many times you pour it back and forth, you can't make the percentage of wine in the two measuring glasses the same.

In the end, it is essentially a question of knowing a[n+1]=a1*an+b1*b[n+1]......1 formula and.

b[n+1]=a2*a[n]+b2*b[n]……2, recursive 2 backwards by a b [n+2] = ......, and then bring in formula 1 to get formula 3, and reversely connect a[n] of formula 2 out, and bring in formula 3 to get such a relation: a[n+2]=k1*a[n+!k2*a[n], which can be solved with a typical eigenequation.

Among them, a1, a2, b1, b2 are self-seeking, and the idea is like this.