Mathematical induction

(1)格式不詳述
For n=2k+1, (因n是奇數), x^2k+1 + y^2k+1=m(x+y), which m is a polynomial
For n=2(k+1)+1, i.e.n=2k+3
x^2k+3 + y^2k+3 =x^2k+1 x^2 + y^2k+1 y^2
=(x^2k+1 + y^2k+1) x^2 - y^2k+1 x^2 + y^2k+1 y^2
=m(x+y)x^2 - y^2k+1 (x^2 - y^2)
=m(x+y)x^2 - y^2k+1 (x-y)(x+y)
=(x+y)[mx^2- y^2k+1 (x-y)]

(2)(a) (k+3)^3=k^3 + 9k^2 + 27k + 27
(b) 格式不詳述
For n=k, k+1, k+2,
k^3 + (k+1)^3 + (k+2)^3 = 9p, which p is a polynomial

For n=k+1, (k+1)+1, (k+1)+2, i.e. n=k+1, k+2, k+3,
(k+1)^3 + (k+2)^3 + (k+3)^3 = (k+1)^3 + (k+2)^3 + (k^3 + 9k^2 + 27k + 27) by2(a)
= [k^3 + (k+1)^3 + (k+2)^3] + 9(k^2 + 3k + 3)
= 9p + 9(k^2 + 3k + 3)
= 9(p + k^2 + 3k + 3)

我只係識中文
1.設p(n)為命題,p(n)=x^n + y^n
當n=1
p(1)=x+y,可被x+y整除
命題p(1)成立
設p(k)為命題,p(k)=x^k+y^k=m(x+y)
x^k=m(x+y)-y^k
當n=k+1
p(k+1)=x^(k+1)+y^(k+1)
=x x^k+y y^k
=x [m(x+y)-y^k]+y y^k
=mx(x+y)-x*y^k+y y^k
=mx(x+y)-y^k(x+y)
=(x+y)[mx-y^k],可被x+y整除
根據.............,命題p(n)成立

2.a (k+3)^3
=k^3+3k^2+3k+1
b.我唔明英文,唔知點計,sor