# Sum notation troubles

1. Oct 14, 2009

### Mentallic

1. The problem statement, all variables and given/known data
1) Express 8+27+64+.... with n terms as a sum notation
2) Expand $(1+x)^n(2+x)^n$ in sum notation

3. The attempt at a solution
I know this is very simple but I've simply forgotten.

1) Ok so firstly I am assuming the sum is meant to be $2^3+3^3+4^3+...+n^3$

So, I think the sum is supposed to be expressed like this:

$$\sum ^{n}_{k=2} k^3$$

But from every other sum I've seen, they all start with k=0 so is it necessary to have the limit starting at 0?

$$\sum ^{n-2}_{k=0} (k+2)^3$$

2) For $(1+x)^n(2+x)^n$ I will express each factor as a sum, and multiply them together.

$$\left[ \sum^{n}_{r=0} ^{n}C_{r}.x^r\right]\left[ \sum^{N}_{R=0} ^{N}C_{R}.2^{N-R}.x^R\right]$$

Now, is it correct if I merged both factors as such?:

$$\ \sum^{n}_{r=0}\sum^{N}_{R=0}^{n}C_{r}.^{N}C_{R}.2^{N-R}.x^{r+R}$$

2. Oct 14, 2009

### Dick

No, it's not necessary to start a sum at zero. You can express it either way. And there is nothing wrong with your merged sum either.

3. Oct 14, 2009

### Mentallic

Thanks for clarifying that for me.

So then if I see a merged sum like I've done, then I can imagine them as being multiple sums (I can go backwards and seperate each sum into 2 factors)?

And I'm having trouble imagining how to expand each sum into a long expression. For small n,N>0, say, n=3, N=2, I'm unsure how to be expanding the sum. Of course, I wouldn't even attempt it for larger n and N since I know how long and crazy it can get.

4. Oct 14, 2009

### Dick

If you have a double sum and you can separate the thing you are summing over into two factors, each one of which is dependent only on one of the summation indices then, sure you can split it back into a product of two single sums. Is that what you mean? You are getting a little abstract here.

5. Oct 14, 2009

### Mentallic

Yes sorry that is what I meant.

But lets say for a second that I couldn't convert a double sum back into a product of 2 single sums simply, how would I expand, say,

$$\ \sum^{3}_{r=0}\sum^{2}_{R=0}^{3}C_{r}.^{2}C_{R}.2^ {2-R}.x^{r+R}$$

Of course I could go backwards and convert this back into a product of 2 factors: $$(1+x)^3(2+x)^2$$ and expand each factor and then multiply the 2 factors as normal, but what if I didn't know how to get back to this point and had to expand the double sum above?

6. Oct 15, 2009

### Staff: Mentor

For the first problem it says "Express 8+27+64+.... with n terms as a sum notation" (underline added).
Your sum has n - 1 terms:
$$\sum ^{n}_{k=2} k^3$$

The number of terms is the ending index - starting index + 1, which is n - 2 + 1 = n - 1 for your sum. As Dick pointed out, there's no rule that says the beginning index has to be 0, so to get exactly n terms in your sum, adjust the ending index.

7. Oct 15, 2009

### Mentallic

Oh thanks for spotting that Mark. I didn't think about getting n terms exactly, I just threw in the first term 2 and the last term as an n.
ok so the sum notation should be:

$$\sum^{n+1}_{k=2}k^3$$

Now, just an answer to post #5 is still pending. If anyone can help me to understand what I'm looking for when expanding a double sum like this, it would be greatly appreciated.

8. Oct 15, 2009

### Dick

For each value of r from 0 to 2 you would expand the remaining single sum for R from 0 to 3 into four terms and then add them all together getting twelve terms.

9. Oct 15, 2009

### Mentallic

Ahh thanks that makes plenty of sense.
Ok I'm going to try expand it, I hope I get this right

$$\ \sum^{3}_{r=0}\sum^{2}_{R=0}^{3}C_{r}.^{2}C_{R}.2^ {2-R}.x^{r+R}$$

$$=\sum^2_{R=0}^3C_0.^2C_R.2^{2-R}.x^{0+R}$$
$$+\sum^2_{R=0}^3C_1.^2C_R.2^{2-R}.x^{1+R}$$
$$+\sum^2_{R=0}^3C_2.^2C_R.2^{2-R}.x^{2+R}$$
$$+\sum^2_{R=0}^3C_3.^2C_R.2^{2-R}.x^{3+R}$$

$$= ^3C_0.^2C_0.2^{2-0}.x^{0+0}+^3C_0.^2C_1.2^{2-1}.x^{0+1}+^3C_0.^2C_2.2^{2-2}.x^{0+2}$$
$$+^3C_1.^2C_0.2^{2-0}.x^{1+0}+^3C_1.^2C_1.2^{2-1}.x^{1+1}+^3C_1.^2C_2.2^{2-2}.x^{1+2}$$
$$+^3C_2.^2C_0.2^{2-0}.x^{2+0}+^3C_2.^2C_1.2^{2-1}.x^{2+1}+^3C_2.^2C_2.2^{2-2}.x^{2+2}$$
$$+^3C_3.^2C_0.2^{2-0}.x^{3+0}+^3C_3.^2C_1.2^{2-1}.x^{3+1}+^3C_3.^2C_2.2^{2-2}.x^{3+2}$$

$$=4+4x+x^2$$
$$+12x+12x^2+3x^3$$
$$+12x^2+12x^3+3x^4$$
$$+4x^3+4x^4+x^5$$

$$=x^5+7x^4+19x^3+25x^2+16x+4$$

Test: x=2

$$LHS=(1+2)^3(2+2)^2=3^3.4^2=432$$
$$RHS=(2)^5+7(2)^4+19(2)^3+25(2)^2+16(2)+4=32+112+152+100+32+4=432=LHS$$

Seems right
Thanks for the help, I have officially mastered sums notation hehe

10. Oct 16, 2009

### Dick

SEEMS right? It IS right. (1+x)^2*(2+x)^3=x^5+7x^4+19x^3+25x^2+16x+4. You could just multiply it out. No need to put a number in to test. But I guess convincing yourself of something by doing it the HARDEST way possible and verifying it is convincing. I hope. I guess?

Last edited: Oct 16, 2009
11. Oct 16, 2009

### Mentallic

I wanted to avoid expanding it out, as I could've just done that in the first place! And imagine if it was much larger n,N (ok the sum would be ridiculous too but thats besides the point). Testing for confirmation wouldn't prove it, but supports it beyond reasonable doubt and also, it's not the hardest way! Just a hard way :tongue2:

Now that I look at it again... wtf was I thinking? haha thanks for the constructive criticism