First show the statement holds for [tex]n=1[/tex]. The left hand side is just 8, and the right hand side is [tex]4(1)(1+1)=8[/tex], so it's true.
Assume the statement holds for [tex]n=k[/tex], i.e.
[tex]8+16+\cdots+8(k-1)+8k=4k(k+1)[/tex]
and use this to show it holds for [tex]n=k+1[/tex], i.e.
[tex]8+16+\cdots+8(k-1)+8k+8(k+1)=4(k+1)(k+2)[/tex]
By the assumption above, you have
[tex]\underbrace{8+16+\cdots+8(k-1)+8k}_{n=k}+8(k+1)=4k(k+1)+8(k+1)=4(k+1)(k+2)[/tex]
so the statement is true for all [tex]n\ge1[/tex].
