# Change of summation

1. Aug 14, 2016

### kwal0203

1. The problem statement, all variables and given/known data

Using change of summation index show that:

$$\sum_{k=1}^{n} (k + 1)^3 - \sum_{k=1}^{n} (k-1)^3 = (n + 1)^3 + n^3 - 1$$

Hence show that:

$$\sum_{k=0}^{n} k^2 = \frac{n}{6} (n + 1)(2n + 1)$$

2. The attempt at a solution

For the first part I changed the summation index like this:

$$\sum_{k=2}^{n+1} k^3 - \sum_{k=0}^{n-1} k^3 = (n + 1)^3 + n^3 - 1$$

Clearly when you get rid of the terms that are common to both summations you are left with the right hand side of the equation.

For the second part, I can prove it by induction but don't see how it's related to the first part of the question.

So far I've done this:

$$\sum_{k=1}^{n} (k + 1)^3 - \sum_{k=1}^{n} (k-1)^3 = \sum_{k=1}^{n} [(k + 1)^3 - (k-1)^3] = \sum_{k=1}^{n} 6k^2 + 2$$

But not really sure where I'm going with this.

Any help appreciated.

2. Aug 14, 2016

### blue_leaf77

$$\sum_{k=1}^{n} (6k^2 + 2) = (n + 1)^3 + n^3 - 1$$
then solve for $\sum_{k=1}^{n} k^2$.

3. Aug 14, 2016

### kwal0203

Ah of course.

Thanks for that.