# Index shifting in summation FORMULAS

1. Mar 13, 2010

### Xyius

So I am trying to derive a formula from one of the standard summation formulas except starting at a different index. So if I have the series..

$$\sum i = \frac{n(n+1)}{2}$$

Where "i" runs from 1 to n. (I dont know how to put it in the code.) If I want to make the series start from zero, I would use another index equal to, "k=i-1" And now my upper limit will be "n-1" So now I can replace "i" in the sum with "k+1", Shouldn't the new formula for the partial sums be..

$$\sum k+1 = \frac{(n-1)(n)}{2} + (n-1)$$
Problem is, it isnt. A simple test will prove this. Whats wrong?

2. Mar 13, 2010

### jbunniii

Your new formula is wrong. (By the way, you can click on the equations below to see the Latex code for putting the indices on the sum.)

The original sum is equivalent to

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

Note that the sum starts with $k=0$, not $k=1$.

This can be rewritten as

$$\sum_{k=0}^{n-1}k + \sum_{k=0}^{n-1} 1$$

The first sum is the same whether it starts with $k=0$ or $k=1$, but the second sum is not.

Last edited: Mar 13, 2010
3. Mar 14, 2010

### HallsofIvy

Staff Emeritus
You can also change the "starting point" by changing the index.

Given
$$\sum_{i= 0}^{n-1} i$$

Let k= i+1 so that when i= 0, k= 0+1= 1 so the lower limit is i= 1. Then when i= n-1, k= (n-1)+ 1= 1 so the upper limit is k= n. Finally, of course, since k= i+ 1, i= k- 1. There's your error- your sum should be of k- 1, not k+ 1:
$$\sum_{k=1}^n k-1$$

4. Mar 14, 2010

### Xyius

Hmm I think I still have some issues. Since the original formula for the partial sums is..

$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$

Since "n" represents the upper limit, the new formula for the series I am after should be just replacing all the "n" with "n-1."
$$\sum_{k=0}^{n-1} k+1 =\frac{(n-1)(n)}{2} + \sum_{k=0}^{n-1} 1$$

Now you say up until this part is correct? For the second sum, shouldn't the sum of this be just the upper limit? For example..

$$\sum_{i=1}^{3} 1 = 3(1) = 3$$

Shouldn't the same thing apply for this sum? Shouldn't it therefore be..
$$\sum_{k=0}^{n-1} 1 = 1(n-1) = n-1$$

Or should it equal "n" because the index is zero?

5. Mar 14, 2010

### jbunniii

Actually both the lower and the upper limit have changed, right? You are now summing from 0 to n-1, whereas the original formula sums from 1 to n.

Now it happens that

$$\sum_{k=0}^{n-1}k = \sum_{k=1}^{n-1}k$$

i.e. it doesn't matter whether you start summing from $k=0$ or $k=1$ in THIS PARTICULAR SUM. That is because the summand is $k$, so the $k=0$ term is 0: it has no effect on the value of the sum. You can therefore start summing from $k=1$ and get the same answer.

Yes, it's correct up to this point.

The summand is $1$, and you are summing it $n$ times. The $k=0$ term DOES matter in this case, because that term is not zero.

Therefore

$$\sum_{k=0}^{n-1} 1 = n$$

Think of it as having $n$ marbles with labels on them. It doesn't matter if the labels read $0$ through $n-1$ or $1$ through $n$, there are still $n$ marbles.

I could just as well write

$$\sum_{k=1000}^{1000+n-1} 1$$

and the sum would still be $n$.

Last edited: Mar 14, 2010
6. Mar 14, 2010

### Xyius

Makes complete sense! Thanks a lot!

So if I wanted to derive a formula for the squared sum. The normal formula is.

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

So therefore the new sum should be..

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

Would this be correct?

7. Mar 14, 2010

### jbunniii

Yes, it looks right to me. And I tried plugging in n = 20 and verified that the two formulas give the same answer. Of course they're really the same formula, in the sense that if you carry out the addition of the three terms in your expression by putting them over a common denominator and simplifying, you should end up with the original formula.

8. Mar 14, 2010

### Xyius

Thanks a lot man! :D!!