Not understanding this series representation

[Kyle Nemeth]

[mentor note: thread moved from non-hw forum to here hence no homework template]

Can someone explain to me how it is that

$$\sum_{n=a}^b (2n+1)=(b+1)^2-a^2$$

I thought it would be $$\sum_{n=a}^b (2n+1)=(2a+1)+(2b+1)$$

but I am clearly very wrong. I would greatly appreciate any help.

 

[Charles Link]

It is summing the series from a to b (integers). This basically means you can take the difference of the sum from (n=1 to n= b) and of the sum from (n=1 to n=a-1). Both of those sums you should be able to compute. $\\$ Note: I think problems of this kind normally would go to the homework section. Please post in the homework section if it involves a homework exercise. :-) :-)

 

[jedishrfu]

The 2n+1 should give you a clue especially if a=0 then you get the sum 1+3+5+...

We can't solve this problem for you so please show some work before we post more.

 

[dkotschessaa]

Wow, took me a bit to figure out what everyone was saying! I thought combinatorics prepared me for any summation I might see!

Kyle,
I don't know if you're missing some qualifiers or something from your statement. But it would have been clearer to me if it was written:
$$\sum_{n=a}^b (2n+1)=(b+1)^2-a^2 , \qquad \forall a,b \in \mathbb{Z} a \leq b $$

Think of it as an identity that helps you compute a summation for two integers meeting that criteria, rather than a formula to just write out. To see that it works, plug in a couple of numbers.

It's a nice identity and I've either never seen it, or it's so basic I have forgotten.

-Dave K

 

[jedishrfu]

Wow, took me a bit to figure out what everyone was saying! I thought combinatorics prepared me for any summation I might see!

Kyle,
I don't know if you're missing some qualifiers or something from your statement. But it would have been clearer to me if it was written:
$$\sum_{n=a}^b (2n+1)=(b+1)^2-a^2 , \qquad \forall a,b \in \mathbb{Z} a \leq b $$

You're definitely a true mathematician now! :-)

Yes, its important to indicate the membership bounds of a and b to make this problem solvable.

 

[Kyle Nemeth]

I apologize for the sloppy post. Thanks everyone for the help and guidance.

 

[jedishrfu]

Your work? Mathematicians are standing by to answer your questions... :-)

 

[dkotschessaa]

You're definitely a true mathematician now! :-)

If that's true I should probably go and prove it now...

I apologize for the sloppy post. Thanks everyone for the help and guidance.

Was it written with the bounds like that? Just curious, might not have been you.

I learned something anyway.

-Dave K

 

[Logical Dog]

I believe, it has to do with the formula of an arithmetic progression, and mathematical induction (how to arrvie at the identity). Combining them or something. But mathematical induction only applies in the set of natural numbers. and also I don't see how the identity applies to negative integers, as the squaring removes the negative one one side but not the other.

edit: oh man, I am really missing something here