Using the binomial formula in the form $(k - k)^n$

Question

While proving a certain property of the number theoretic mobius function, namely that it is invertible in the monoid of multiplicative functions and its inverse is the constant function $\eta\left(n\right) = 1$,

I used an argument which boils down to an expression of the form $$ \sum_{j = 0}^{n} \binom{n}{j}(-1)^{j} $$ which I then claim is equal to $\underline{zero}$ since this is a special case of the Binomial Theorem $$ \left(x + y\right)^{n}\quad \mbox{when}\quad x = 1\ \mbox{and}\ y = -1 $$ and hence $\left(x + y\right){^n} = \left(1 - 1\right)^{n} = 0$.

This is kosher right? Nothing prevents me from using the binomial theorem under these conditions?

Absolutely. Good argument. For a combinatorial proof of the same identity, see this question — lulu, Jun 30 '24 at 19:56
This is also the observation that the sum of the entries (alternating the signs) in any row of Pascal's triangle (except the 0th row) is 0. — J. Chapman, Jun 30 '24 at 21:16
$$ \sum_{j = 0}^{n}{n \choose j} \left(-1\right)^{j} = \color{red}{\Large\delta_{n0}} $$ — Felix Marin, Jun 30 '24 at 21:59

score 1 · Answer 1 · answered Jul 08 '24 at 19:04

1

Yes, in this case, it works. You can expand $(1-1)^n$ to verify and get $\sum^n_{j=0} (-1)^j \binom{n}{j}$ is equal to $(1-1)^n$.

answered Jul 08 '24 at 19:04

Andy Liu

442

Using the binomial formula in the form $(k - k)^n$

1 Answers1