Step-by-step explanation:

First, that answer is correct. If you note that 2+4+6+...+2n=2∗(1+2+3+...+n)=2n∗(n+1)2=n∗(n+1)2+4+6+...+2n=2∗(1+2+3+...+n)=2n∗(n+1)2=n∗(n+1)

Second. Suppose that it is true, you have 2+4+6+...+2n=n∗(n+1)2+4+6+...+2n=n∗(n+1) This is S(n)S(n) add 2(n+1) to both sides, giving

2+4+...+2n+2(n+1)=n∗(n+1)+2(n+1)=(n+2)(n+1)2+4+...+2n+2(n+1)=n∗(n+1)+2(n+1)=(n+2)(n+1)

which is S(n+1)