Summations: CP1 - Further Maths

Just a quick question, what do you do when r = 0? (the number below the sigma symbol)

Just started year 12.. btw!

The sum starts with r=0. What is the original question?

3 is above the sigma and the sum is (2r+1)

The question says evaluate

So evaluate the sum starting with r=0 and finishing with r=3. There are only 4 expressions to add which you could simply write out.

Yes, I know how to do that. I wanted to ask if there's a way you can turn that into r=1 and solve it using the summation formulas

Sure, what is the r=0 term, then add it onto the (modified) sum starting with r=1. Writing out the terms should make it clear

That's all it said in the q

Just a quick question, what do you do when r = 0? (the number below the sigma symbol)

Just started year 12.. btw!

Hello Jenius,
Welcome to year 12!
Regarding your question, when r = 0, the term inside the sigma notation is undefined. This is because the sigma notation tells us to add up all the terms from r = 1 to n, and 0 is not included in this range.

So, when r = 0, we simply ignore the term
Thank you

As per the previous post add the zeroth term
2*0 + 1 = ...
onto the result of the normal sum which starts at r=1. Simple as that as the sum is
zeroth term + first term + ... + third term

Ohhh. I get it now, thanks! 💯

Hello Jenius,
Welcome to year 12!
Regarding your question, when r = 0, the term inside the sigma notation is undefined. This is because the sigma notation tells us to add up all the terms from r = 1 to n, and 0 is not included in this range.

So, when r = 0, we simply ignore the term
Thank you

Thats simply wrong.

O.K. I see. About your previous comment, just asking, can you use the summation formula 1/2 n(n+1) when r = 0?

For the basic summation of natural numbers fairly obv
0 + 1 + 2 + ... + n = 1 + 2 + .... + n
so yes.

Ok. Thanks You don't need to, but if you want to, can u give me an example? So I can see how it goes

An example of what?

Finding the sum of something when r=0 using the summation formula 1/2 n ( n + 1 )