Expressing Sum Notations for 8+27+64+... and (1+x)^n(2+x)^n | Solution Steps

[Mentallic]

Homework Statement

1) Express 8+27+64+... with n terms as a sum notation
2) Expand (1+x)^n(2+x)^n in sum notation

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}

 

[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.

 

[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 separate 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.

 

[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.

 

[Mentallic]

Yes sorry that is what I meant.

But let's 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?

 

[Mark44]

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.

 

[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.

 

[Dick]

Yes sorry that is what I meant.

But let's 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?

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.

 

[Mentallic]

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.

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

 

[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 into test. But I guess convincing yourself of something by doing it the HARDEST way possible and verifying it is convincing. I hope. I guess?

 

[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 that's 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

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