Prove $\sum_{r=0}^n 6r=3n(n+1)$ using induction

Question

Prove$$\sum_{r=0}^n 6r=3n(n+1)$$using Induction

I'm a little confused as to how I would calculate the latter

Answer 1

Assume that it's true For $n=k$

$$\sum_{r=0}^k 6r=3k(k+1)$$

Then show that it's also true for n=(k+1)

$$\sum_{r=0}^{k+1} 6r=6(k+1)+\sum_{r=0}^{k} 6r=6(k+1)+3k(k+1)=3k^2+9k+6=3(k+1)(k+2)$$