Question

prove by mathematical induction that for n even n^2+ 2n is a multiple of 8

Answer 1

I did it till this step (k+1) (k+3), what to do after this?

Answer 2

P(2)=8 == 8*1 ==> P(2) is true
Let P(n) be true
==> n^2+2n=8k
P(n+2)=(n+2)^2+2*(n+2)
= n^2+2n+4n+4+4
=8k+4n+8
=8(k+1)+4n
n is even ==>n= 2x
=8(k+1)+4*2x
=8*(k+1+x)
==>P(n+2) is a multiple of 8
Hence by induction P(n)= n^2+2n is a multiple of 8 for n even

Answer 3

Let n=2x then,
n^2+2n =4x^2 + 4x = 4*x*(x+1)
Now, product of any two consecutive integer is always even So,
4*x*(x+1) is always divisible by 8...
Thus, n^2+2n is a multiple of 8 if n is even.