How can you prove this discrete math induction statement?

[iPetey]

Homework Statement

Homework Equations

base case: n=1

The Attempt at a Solution

im not sure where to start because the examples that my professor showed us did not have a n(n-1) (n+1) but rather (p+1)P=1+1)(2(p+1)+1)

im just very lost in this example

 

[SEngstrom]

Assume the statement is true for n=m. Add one more term and see if the statement is still true for n=m+1. If it is and also true for n=2, then true for all n...

 

[Outlined]

you do this kind of stuff by induction

 

[iPetey]

i know its done by induction but i don't know the proper steps to do it.

 

[SEngstrom]

Assume that the statement you have is true. Now add another term (do the sum to N instead of N-1). Add that term to the RHS and see if you can reshape that expression to the same form as it is now, only with N*(N+1)*(N+2)/3 instead. If you can, it must hold true for all N from there on. Then it remains to show that it holds for a sum with just one term.

 

[eumyang]

Assume that the statement you have is true. Now add another term (do the sum to N instead of N-1). Add that term to the RHS and see if you can reshape that expression to the same form as it is now, only with N*(N+1)*(N+2)/3 instead. If you can, it must hold true for all N from there on. Then it remains to show that it holds for a sum with just one term.

Personally I don't like re-using the variable N; instead I would use k:
Assume true for n = k:
$$\displaystyle \sum_{i=1}^{k-1} i(i +1) = \frac{k(k-1)(k+1)}{3}$$
Prove true for n = k + 1:
$$\displaystyle \sum_{i=1}^{(k+1)-1} i(i +1) = ... = \frac{(k + 1)(k)(k+2)}{3}$$

OP: a hint would be to "break up" the summation in the n = k + 1 case.