# Spivak Calculus Summation problem

1. May 6, 2015

### Wumbolog

• Member warned about not using the homework template
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 dont understand how he reached his solution and I dont understand how to manipulate these equations "1+....+n"

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2. May 6, 2015

### muscaria

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

3. May 6, 2015

### 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?

4. May 6, 2015

### 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?

5. May 7, 2015

### Wumbolog

You get n amount of n+1? Im sorry I still dont get it

6. May 7, 2015

7. May 7, 2015

### muscaria

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?

8. May 7, 2015

### muscaria

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?

9. May 7, 2015

### certainly

What problem did you have ?

10. May 7, 2015

### SammyS

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

11. May 7, 2015

### muscaria

Duly noted.

12. May 8, 2015

### Wumbolog

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 im supposed to add to get zero -2(1+...+n) is still there

I dont understand

13. May 8, 2015

### SammyS

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

14. May 8, 2015

### Wumbolog

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

15. May 8, 2015

### SammyS

Staff Emeritus
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. $\displaystyle\ \sum_{i=1}^{2n}i\ \ ?$
.

16. May 9, 2015

### Wumbolog

17. May 9, 2015

### SammyS

Staff Emeritus

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?