Binomial Expansion alternating Sum

[Punkyc7]

use the binomial expansion formula to find the alternating sum of the numbers in row n of Pascals triangle.
nC0-nC1+nC2...


So I wrote a few rows of the triangle and it looks like once you get past row zero the alternating sum of the numbers in the rows all add to zero.

Does that work for an answer for this problem or does there have to be some type of proof... if it has to be a proof would you have to use induction?

 

[dynamicsolo]

What is the general formula for calculating the terms in the nth row of "Pascal's triangle"* ? Is there any special symmetry among the terms? Is there any general way to express pairing up terms of opposite sign?

*actually discovered outside of Europe many centuries earlier

 

[Punkyc7]

you add the two terms above the number you want in the triangle.
Yes there is
not sure what you mean by the third question

 

[vela]

Try looking at (a+b)ⁿ and determine what values of a and b will give you the sum you're trying to calculate.

 

[Ray Vickson]

You are ignoring what the question asked you to do, namely: *use the binomial expansion*.

RGV

 

[SammyS]

Consider (1 + (-1))ⁿ .

 

[Punkyc7]

that would get you 0

 

[Punkyc7]

You are ignoring what the question asked you to do, namely: *use the binomial expansion*.

RGV

So how do you do that beside using our triangle?

 

[process91]

that would get you 0

Instead of driving right to trying to simplify the expression $(1+(-1))^n$, what happens when you apply the binomial theorem to it? Now what happens when you solve it?

 

[Punkyc7]

you get

nC0 (1)^n (-1)^0+ nC1 (1)^n-1 (-1)^n ...+nCn (1)^0 (-1)^n

right?

everything should cancel everything out

 

[process91]

you get

nC0 (1)^n (-1)^0+ nC1 (1)^n-1 (-1)^n ...+nCn (1)^0 (-1)^n

right?

everything should cancel everything out

What you want to do is think about it two ways. When you apply the Binomial Theorem, the numbers you get should trigger something familiar related to the question. When you just solve it straight off, you saw that it equals zero. I think you've got it, but you need to just connect the pieces. Perhaps try applying the binomial theorem for some small real values of n.

 

[Punkyc7]

if the row number is odd then there is even number of terms that nicely cancel out and if the row number is even there is an odd number of terms that when you work out the sums cancel out, but how does that show the sum is zero. I know that it is, can you just say clearly the sum of any row greater than zero is zero. I don't understand how you can just say that if you don't have any numbers to work with.

 

[vela]

you get

nC0 (1)^n (-1)^0+ nC1 (1)^n-1 (-1)^n ...+nCn (1)^0 (-1)^n

right?

Not exactly. The bolded exponent is wrong.

Let n=3. What do you get for the expansion? What about when n=4? Do you see the pattern? How do they compare what you're asked to find?

 

[process91]

if the row number is odd then there is even number of terms that nicely cancel out and if the row number is even there is an odd number of terms that when you work out the sums cancel out, but how does that show the sum is zero. I know that it is, can you just say clearly the sum of any row greater than zero is zero. I don't understand how you can just say that if you don't have any numbers to work with.

Sure you do, you said it yourself when you were presented with $(1+(-1))^n$; your immediate reply was that it was equal to zero. I feel like I'm giving it too much away, so I'll just say this - you start like this:

$(1+(-1))^n=\sum_{k=0}^n {n \choose k} 1^k (-1)^{n-k}$

You should be able to write in one line below this your answer.

 

[Punkyc7]

yeah that was a mistake I should have caught that when i was typing it

1331
14641if n is odd everything cancels out nicely 1 cancel out -1 3 cancel outs -3
if n is even you have to sum everything but how do you say that when you have n things...

 

[vela]

yeah that was a mistake I should have caught that when i was typing it

1331
14641

What happened to the plus and minus signs?

if n is odd everything cancels out nicely 1 cancel out -1 3 cancel outs -3
if n is even you have to sum everything but how do you say that when you have n things...

 

[Punkyc7]

1 -3 3 -1
1 -4 6 -4 1

I thought you were just looking for the numbers but there they are

 

[vela]

I'm asking you to write what

nC0 (1)^n (-1)^0 + nC1 (1)^n-1 (-1)^1 + … + nCn (1)^0 (-1)^n

equals when n=3 and n=4, or more generally to see what it's equal to in the general case. It's not equal to the sequences of numbers you wrote above. It's equal to the sum of those numbers:

1 + (-3) + 3 + (-1) = 1 - 3 + 3 - 1
1 + (-4) + 6 + (-4) + 1 = 1 - 4 + 6 - 4 + 1

So what do you get for the general case?

 

[Punkyc7]

nC0-nC1 +nC2...

but how do you know which sign the last term is going to be because it depends on n. Also without actual numbers how can you say that everything just cancels out.

 

[vela]

So that's the sum you wanted to compute. The sign of the last term will depend on n. It'll be (-1)ⁿ. Now, from what expression did you come up with

nC0 (1)^n (-1)^0 + nC1 (1)^n-1 (-1)^1 + … + nCn (1)^0 (-1)^n

in the first place, and what was it equal to?

 

[Punkyc7]

I just assumed we were still using( 1 +(-1))^n which is equal to 0.

 

[vela]

Exactly! So you just need to put the pieces together.

 

[Punkyc7]

So would you say something like

The sum of the nth row of pascals triangle is
nC0 1^n (-1)^0 + nC1 1^n-1 (-1)^1 +...+ nCn 1^0 (-1)^n =(1+ (-1))^n =0
QEDshould I mention what happens when n is 0 ?

 

[vela]

Yup, that's pretty much it. And yes, if the problem didn't explicitly exclude the n=0 case, you should discuss it as well.

 

[Punkyc7]

Ok thank you so much. I have a hard time with abstract math, it just doesn't seem to click with me.