Parrot Guy's question at Yahoo Answers regarding a summation proof by induction

[MarkFL]

Here is the question:

Mathematical Induction Problem help?



Use Mathematical Induction to prove the following statement:

2 + 10 + 24 + 44 + . . . + n(3n - 1) = n^2(n+1)

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Parrot Guy,

We are given to prove:

\(\displaystyle \sum_{j=1}^n\left(j(3j-1) \right)=n^2(n+1)\)

First, we check to see if the base case $P_1$ is true:

\(\displaystyle \sum_{j=1}^1\left(j(3j-1) \right)=1^2(1+1)\)

\(\displaystyle 1(3\cdot1-1)=1(1+1)\)

\(\displaystyle 2=2\)

The base case is true, so next we state the induction hypothesis $P_k$:

\(\displaystyle \sum_{j=1}^k\left(j(3j-1) \right)=k^2(k+1)\)

As our induction step, we may add \(\displaystyle (k+1)(3(k+1)-1)\) to both sides:

\(\displaystyle \sum_{j=1}^k\left(j(3j-1) \right)+(k+1)(3(k+1)-1)=k^2(k+1)+(k+1)(3(k+1)-1)\)

On the left, incorporate the new term into the summation and on the right, factor and distribute:

\(\displaystyle \sum_{j=1}^{k+1}\left(j(3j-1) \right)=(k+1)\left(k^2+3k+2 \right)\)

Factor further on the right:

\(\displaystyle \sum_{j=1}^{k+1}\left(j(3j-1) \right)=(k+1)(k+1)(k+2)\)

Rewrite the right side:

\(\displaystyle \sum_{j=1}^{k+1}\left(j(3j-1) \right)=(k+1)^2((k+1)+1)\)

We have derived $P_{k+1}$ from $P_{k}$ thereby completing the proof by induction.