A. maths M.I. Question ERGENT HW!!!!!!!!!!!!!!!!!!!!!!! 20pt

Let P(n)=x^n-nxy^(n-1)+(n-1)y^n

for n=2

x^2-2xy+y^2=(x-y)^2

therefore P(2) is divisible by (x-y)

Assume P(k) is divisible by (x-y)^2

hence we have,

x^k - kxy^(k-1) + (k-1)y^k = M(x-y)^2 -----(1)

=>x^k=M(x-y)^2+kxy^(k-1)-(k-1)y^k

for n=k+1

P(k+1)=x^(k+1)-(k+1)xy^k+ky^(k+1)

by using the (1), we have

P(k+1)=x[M(x-y)^2+kxy^(k-1) - (k-1)y^k] - (k+1)xy^k+ky^(k+1)

=Mx(x-y)^2 + kx^2y^(k-1) - (k-1)xy^k - (k+1)xy^k+ky^(k+1)

=Mx(x-y)^2 +ky^(k-1)[x^2+y^2] - xy^k[2k]

=Mx(x-y)^2 +ky^(k-1)[x^2-xy+y^2]

=Mx(x-y)^2+ky^(k-1)(x-y)^2
=(x-y)^2[Mx+ky^(k-1)]
therefore P(k+1) is divisible by (x-y)^2
hence, by MI, P(n) is divisible by (x-2)^2 for all integers n≥2




2007-10-26 09:29:39 補充：
i)4*9^n-(5n 4)*4^n is divisible by 25sub x=9, y=5 into P(n), we haveP(n)=9^n-(n)(9)(4)^(n-1) (n-1)4^n=9^n-4^(n-1)[9n-4(n-1)]=9^n-4^(n-1)[5n 4]multiply it by 4, we have4*9^n-(5n 4)*4^n=4P(n)therefore 4*9^n-(5n 4)*4^n is divisible by 25 (since (9-4)^2=25)

2007-10-26 09:38:54 補充：
ii)7^n-(3n-4)*4^(n-1) is divisible by 9 is there a problem about the questionI think it should be :ii)7^n-(3n 4)*4^(n-1) is divisible by 9 if my guess is right, then just by putting x=7, y=4 into P(n) then, you can get the result.

2007-10-26 09:39:44 補充：
I think it should be :ii)7^n-(3n＋4)*4^(n-1) is divisible by 9

use cauculater