Mathematical induction555

Prove, by mathematical induction, that n(n+1)(n+2)(3n+5) is divisible by 24 for all positive integers n.

Solution:

When n=1 , (1)(1+1)(1+2)(3+5)=48=24x2 , it is true for n=1 ,

Assume it is true for n =k, where k is a positive integer,

i.e. k(k+1)(k+2)(3k+5)=24 M , where M is a positive integer ,

Consider n=k+1 ,

(k+1)(k+2)(k+3)(3k+8)

=(3k+5)(k+1)(k+2)(k+3)+3(k+1)(k+2)(k+3)

=24M+3(k+1)(k+2)(k+3)

( there are three continuous number a,a+1 and a+2 , then there must be a number is multipute of 2 and 3 , i.e. 6N)

=24M+24N=24(M+N) , it is true for n=k+1

By mathematical induction , it is true for all positive integer n.

Prove, by mathematical induction, that x^(n+2) + (x+1)^(2n+1) is divisible by (x^2) + x + 1 for all positive integers n.

When n=1 , x^3+(x+1)^3

=(x+x+1)[(x+1)^2-x(x+1)+x^2] [by a^3+b^3=(a+b)(a^2-ab+b^2)]

=(2x+1)(x^2+2x+1-x^2-x+x^2)

=(2x+1)(x^2+x+1)

so it is true for n=1 .

Assume it is true for n = k , where k is a positive integer ,

i.e. x^(k+2)+(x+1)^(2k+1) = (x^2+x+1) p (x) , where p(x) is a polynomial with real coefficients.

Consider n=k+1 ,

x^(k+3)+(x+1)^(2k+3)

= [(x^2+x+1)p(x)]-(x+1)^(2k+1)]x+(x+1)^(2k+3)

=x(x^2+x+1)p(x)-x(x+1)^(2k+1)+(x+1)^(2k+3)

=x(x^2+x+1)p(x)+(x+1)^(2k+1)[(x+1)^2-x]

=x(x^2+x+1)p(x)+(x+1)^(2k+1)[x^2+x+1]

=(x^2+x+1)[x p(x)+(x+1)^(2k+1) ]

so it is true for n=k+1 ,

By mathematical induction , it is true for all positive integer n.