# Simple series confusion

1. Jun 6, 2012

### synkk

I understand that they have done $$\displaystyle\sum_{r=k-1}^{2k} r =\sum_{r=1}^{2k}r - \sum_{r=1}^{k-1-1}r$$

But where did r = 1 come from as r = k-1 before is what I don't understand.

Also this question:

Show that $$\displaystyle\sum_{p=3}^{n}(4p+5) = (2n+11)(n-2)$$
$$\displaystyle4\sum_{p=1}^{n}p + 5\sum_{p=1}^{n}1 = ...$$

How do you go from p=3 to p=1

2. Jun 6, 2012

### Infinitum

I'll break this down to simple numbers, but the same logic applies for your problem.

Lets say you are asked for the sum $3+4+5$. Doing this in summation form, you could add the numbers from 1 to 2 i.e $1+2$ and subtract it from the sum of 1 to 5, i.e $1+2+3+4+5$, which gives you the same result.

Other way of seeing this is, say you are asked to find the sum $S$. You introduce another quantity $Q$ such that you know the value of $S+Q$ and $Q$. Subtracting these known values, you can find $S$.

3. Jun 6, 2012

### synkk

I can see how it applies to the first question, but not second going from p=3 to p=1 makes no sense to me.

4. Jun 6, 2012

### Infinitum

Well, the second question is wrong. Or you might not have read it properly.

That's why I didn't include it in my quote :tongue:

5. Jun 6, 2012

@ Infinitum, wait, really? Because Wolfram Alpha and I both got the provided answer...

@ synkk: $\displaystyle \sum_{p=1}^{n} 4p + 5 = \displaystyle \sum_{p=1}^{2} 4p + 5 + \displaystyle \sum_{p=3}^{n} 4p+5$

6. Jun 6, 2012

### synkk

got it guys thanks

7. Jun 7, 2012

### Infinitum

The summation you are trying to prove is of course correct. What's wrong is....

$$\displaystyle\sum_{p=3}^{n}(4p+5) = \displaystyle4\sum_{p=1}^{n}p + 5\sum_{p=1}^{n}1$$