Spivak Calculus Summation problem

[Wumbolog]

Hi, I've enclosed my problem and attempt at solution below. I had problems with the latex so I photographed a picture of my work. The first top half is my attempt at the solution and the bottom is the solution that Spivak provides.

I don't understand how he reached his solution and I don't understand how to manipulate these equations "1+...+n"

Your feedback is much appreciated.

 

[muscaria]

Do you understand how he got from the first to the second line?

 

[muscaria]

for 1+...+n, let's take n=11 as an example: If you think of 1+2+3+...+11, that is = (11+1) + (10+2) + (9+3) + ... + (6+6) and these terms are all 12 or (n+1). Now, how many these (n+1) terms do you get?

 

[muscaria]

Once you have figured that out, then look at Spivak's first line: 1 + 3 + 5 + 7 + ...+ (2n - 1)
What can you add to this to make it look like 1 + 2 + 3 + 4 + ...+ (2n - 1) + 2n which you know how to solve?

 

[Wumbolog]

for 1+...+n, let's take n=11 as an example: If you think of 1+2+3+...+11, that is = (11+1) + (10+2) + (9+3) + ... + (6+6) and these terms are all 12 or (n+1). Now, how many these (n+1) terms do you get?

You get n amount of n+1? I am sorry I still don't get it

 

[Wumbolog]

Once you have figured that out, then look at Spivak's first line: 1 + 3 + 5 + 7 + ...+ (2n - 1)
What can you add to this to make it look like 1 + 2 + 3 + 4 + ...+ (2n - 1) + 2n which you know how to solve?

I add even numbers

 

[muscaria]

You get n amount of n+1? I am sorry I still don't get it

Ok maybe I should have explained a bit more what I mean and taken n to be an even number. So imagine you have n numbers where n is even, say n = 10 and you're going to add them all up together. So they're all lined up one after the next 1 2 3 4 5 6 7 8 9 10. Now you could fold this line (or wrap it) so that the 10 is under the 1, the 9 is under the 2, the 8 is under the 3 etc, like:
1 2 3 4 5
10 9 8 7 6

The point of pairing them this way is because as go to the right in the top line you go +1 whereas as you go to the right in the bottom line you go -1 so that the pairs always add to 11. So now how many pairs of 11 do you get?

 

[muscaria]

I add even numbers

Yes precisely you add even numbers to them and now you sum 1+2+3+...+2n and because you added all those even numbers you have to subtract them also (so that you have effectively added 0 to your equation). How do you think Spivak's term -2(1+2+...+n) is related to what I just said?
Does it make sense?

 

[certainly]

I had problems with the latex

What problem did you have ?

 

[SammyS]

Ok maybe I should have explained a bit more what I mean and taken n to be an even number. So imagine you have n numbers where n is even, say n = 10 and you're going to add them all up together. So they're all lined up one after the next 1 2 3 4 5 6 7 8 9 10. Now you could fold this line (or wrap it) so that the 10 is under the 1, the 9 is under the 2, the 8 is under the 3 etc, like:
1 2 3 4 5
10 9 8 7 6

The point of pairing them this way is because as go to the right in the top line you go +1 whereas as you go to the right in the bottom line you go -1 so that the pairs always add to 11. So now how many pairs of 11 do you get?

Yes precisely you add even numbers to them and now you sum 1+2+3+...+2n and because you added all those even numbers you have to subtract them also (so that you have effectively added 0 to your equation). How do you think Spivak's term -2(1+2+...+n) is related to what I just said?
Does it make sense?

Don't do too much of the solution for the OP.

 

[muscaria]

Don't do too much of the solution for the OP.

Duly noted.

 

[Wumbolog]

Yes precisely you add even numbers to them and now you sum 1+2+3+...+2n and because you added all those even numbers you have to subtract them also (so that you have effectively added 0 to your equation). How do you think Spivak's term -2(1+2+...+n) is related to what I just said?
Does it make sense?

So then I have 1+2+3+4+5+(2n-1)+2n

But when I subtract the sum of 2n which is 2(1+...+n)
the second line is 1+2+3+4+5+2n-2(1+...+n), the odd number (2n-1) somehow disappeared and the number I am supposed to add to get zero -2(1+...+n) is still there

I don't understand

 

[SammyS]

So then I have 1+2+3+4+5+(2n-1)+2n

But when I subtract the sum of 2n which is 2(1+...+n)
the second line is 1+2+3+4+5+2n-2(1+...+n), the odd number (2n-1) somehow disappeared and the number I am supposed to add to get zero -2(1+...+n) is still there

I don't understand

What do you mean by "the sum of 2n" ?

 

[Wumbolog]

What do you mean by "the sum of 2n" ?

I put what I meant, which is 2(1+...+n)

 

[SammyS]

So then I have 1+2+3+4+5+(2n-1)+2n

But when I subtract the sum of 2n which is 2(1+...+n)
the second line is 1+2+3+4+5+2n-2(1+...+n), the odd number (2n-1) somehow disappeared and the number I'm supposed to add to get zero -2(1+...+n) is still there

I don't understand

Rather than calling this " the sum of 2n " it's much clearer to say something like, " the sum of all even integers from 2 through 2n. "

In Latex that's $\displaystyle\ \sum_{i=1}^{n}2i\$ .

And you have correctly stated that gives $\displaystyle\ 2\left(\sum_{i=1}^{n}i\right)\$, so that is $\displaystyle\ 2\left(\frac{n(n+1)}{2}\right)$ .

What is the big sum, 1+2+3+4+5+(2n-1)+2n, i.e. $\displaystyle\ \sum_{i=1}^{2n}i\ \ ?$
.

 

[Wumbolog]

Rather than calling this " the sum of 2n " it's much clearer to say something like, " the sum of all even integers from 2 through 2n. "

In Latex that's $\displaystyle\ \sum_{i=1}^{n}2i\$ .

And you have correctly stated that gives $\displaystyle\ 2\left(\sum_{i=1}^{n}i\right)\$, so that is $\displaystyle\ 2\left(\frac{n(n+1)}{2}\right)$ .

What is the big sum, 1+2+3+4+5+(2n-1)+2n, i.e. $\displaystyle\ \sum_{i=1}^{2n}i\ \ ?$
.

I can't figure this out, please help. I need a worked example I can refer to

 

[SammyS]

If you understand the results shown above, I think we can get the correct result using a alternative method to Spivak's.

Now consider the sum of the first n odd natural numbers, $\displaystyle\ \sum_{i=1}^{n}(2i-1)\$.

Split that up into $\displaystyle\ \sum_{i=1}^{n}(2i)-\sum_{i=1}^{n}(1)\\$

What do those last two sums give?