Hello Parrot Guy,
We are given to prove:
$$\sum_{j=1}^n\left(j(3j-1) \right)=n^2(n+1)$$
First, we check to see if the base case $P_1$ is true:
$$\sum_{j=1}^1\left(j(3j-1) \right)=1^2(1+1)$$
$$1(3\cdot1-1)=1(1+1)$$
$$2=2$$
The base case is true, so next we state the induction hypothesis $P_k$:
$$\sum_{j=1}^k\left(j(3j-1) \right)=k^2(k+1)$$
As our induction step, we may add $$(k+1)(3(k+1)-1)$$ to both sides:
$$\sum_{j=1}^k\left(j(3j-1) \right)+(k+1)(3(k+1)-1)=k^2(k+1)+(k+1)(3(k+1)-1)$$
On the left, incorporate the new term into the summation and on the right, factor and distribute:
$$\sum_{j=1}^{k+1}\left(j(3j-1) \right)=(k+1)\left(k^2+3k+2 \right)$$
Factor further on the right:
$$\sum_{j=1}^{k+1}\left(j(3j-1) \right)=(k+1)(k+1)(k+2)$$
Rewrite the right side:
$$\sum_{j=1}^{k+1}\left(j(3j-1) \right)=(k+1)^2((k+1)+1)$$
We have derived $P_{k+1}$ from $P_{k}$ thereby completing the proof by induction.