Proving (a+b)^n = ∑(n_μ)(a^μ)(b^(n-μ)) | a=1, b=-1 using Sequences Test

[AnthonyAcc]

Show that (^{n}_{n}) - (^{n}_{n-1}) + (^{n}_{n-2}) - (^{n}_{n-3}) + ...(^{n}_{0}) = 0

(a+b)^{n} = \sum^{\infty}_{\nu=0} (^{n}_{\nu})a^{\nu}b^{n-\nu}a=1
b=-1

0 = (1+(-1))^{n} = \sum^{\infty}_{\nu=0}(^{n}_{\nu}) 1^{\nu}(-1)^{n-\nu} = \sum^{\infty}_{\nu=0}(^{n}_{\nu})(-1)^{n-\nu} =

...I don't know what to do here...

= \sum^{n}_{\nu=0}(^{n}_{n-\nu})(-1)^{\nu} = (^{n}_{n}) - (^{n}_{n-1}) + (^{n}_{n-2}) - (^{n}_{n-3}) + ...(^{n}_{0})That is if that last equality is correct and makes sense to be there.

Any suggestions?

Also, how can I make it so everything that follows the Sigma doesn't look like it is being superscripted?

Thanks!

 

[Дьявол]

a=1, b=-1
(1+(-1))ⁿ=0

Now try a=-1, b=1 so that

(-1+1)ⁿ=0

Or separate the odds from the even and create two separate sums:
sum (even) - sum (odd) = 0

 

[AnthonyAcc]

Does this \sum^{\infty}_{\nu=0} (^{n}_{\nu}) imply \sum^{n}_{\nu=0} (^{n}_{\nu}) because it's for every n choose \nu so there can only be n many \nus?

 

[Дьявол]

I told you to use something like this:
\sum_{v=0}^{n/2}\binom{n}{n-2v}(-1)^{2v}*1^{n-2v}+\sum_{v=0}^{n/2}\binom{n}{n-2v-1}(-1)^{2v-1}*1^{n-2v-1}

For the both sums a=-1 and b=1, so that 0-0=0

Use \binom{ } { } for binomial coefficients.

Regards.

 

[AnthonyAcc]

A more fundamental question I have, I guess, is if I know that \sum^{\infty}_{\nu=0} (^{n}_{\nu}) a^{\nu} b^{n-\nu} = (a+b)^{n} then do I know that
\sum^{n}_{\nu=0} (^{n}_{n-\nu}) a^{\nu} b^{n-\nu} = (a+b)^{n}? How does the changing of infinity to n and the \nu to n - \nu impact the series?

 

[Dick]

n is an integer, right? Then the generalized binomial coefficient C(n,nu) of n and nu is zero for nu>n, also right? There's a zero factor in the definition. That's how you truncate the infinite sum. Though I'm not sure why you have an infinite sum to begin with. And in that case, sure, C(n,nu)=C(n,n-nu).