How do you simplify summation expressions?

[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

 

[Infinitum]

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.

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.

 

[synkk]

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.

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.

 

[Infinitum]

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.

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

 

[professordad]

@ 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

 

[synkk]

got it guys thanks

 

[Infinitum]

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

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