Htykuut

I would like to receive some help about the next problem.

The problem:

I'm trying to prove the next equation:
$$sum_{k = 0}^{n} frac{(-1)^{-k}}{k!(n - k)!} = 0 quad, n = 1, 2, ...$$

My work until now:
$$sum_{k = 0}^{n} frac{(-1)^{-k}}{k!(n - k)!} = sum_{k = 0}^{n} frac{1}{(-1)^{k}k!(n - k)!} =$$
$$= frac{1}{(-1)^0 cdot n!} + frac{1}{(-1)^1 cdot (n - 1)!} + frac{1}{(-1)^2 cdot 2!(n - 2)!} + frac{1}{(-1)^3 cdot 3!(n - 3)!} + cdot cdot cdot + frac{1}{(-1)^{n - 3} cdot (n - 3)!3!} + frac{1}{(-1)^{n - 2} cdot (n - 2)!2!} + frac{1}{(-1)^{n - 1} cdot (n - 1)!} + frac{1}{(-1)^n cdot n!} quad (1)$$

Because we have $n + 1$ summands i tried to see what happens when $n$ is even or odd number:

1) $n = 2k - 1$, $k in mathbb{N} Longrightarrow$ Now there is $2k$ summands

$$(1) iff
frac{1}{(2k - 1)!}left( frac{1}{(-1)^0} + frac{1}{(-1)^{2k - 1}}right) +
frac{1}{(2k - 2)!}left( frac{1}{(-1)^1} + frac{1}{(-1)^{2k - 2}}right) +
frac{1}{2!(2k - 3)!}left( frac{1}{(-1)^2} + frac{1}{(-1)^{2k - 3}}right) +
cdot cdot cdot +
frac{1}{(k - 1)!k!}left( frac{1}{(-1)^{k - 1}} + frac{1}{(-1)^k}right) = 0$$
This is equal $0$, because the difference in powers of number $-1$ is odd number, so one of them has to be $1$ and another has to be $-1$.

2) $n = 2k$, $k in mathbb{N} Longrightarrow$ Now there is $2k + 1$ summands

$$(1) iff
frac{1}{(2k)!}left( frac{1}{(-1)^0} + frac{1}{(-1)^{2k}}right) +
frac{1}{(2k - 1)!}left( frac{1}{(-1)^1} + frac{1}{(-1)^{2k - 1}}right) +
frac{1}{2!(2k - 2)!}left( frac{1}{(-1)^2} + frac{1}{(-1)^{2k - 2}}right) +
cdot cdot cdot +
frac{1}{(k - 1)!(k + 1)!}left( frac{1}{(-1)^{k - 1}} + frac{1}{(-1)^{k + 1}}right) +
frac{1}{(-1)^k cdot k!k!} quad Longrightarrow quad ?$$

Here, i can't use the same princip as in the first case. Also, the last summand is the $k + 1$ -st summand, he is in the middle of this sum and doesn't have a pair.

My question:

Please, could you help me with some advice or a hint that i could use to finnish my proof of the given equation?

Thank you, for your time and your help!

HINT

Actually you were thinking a bit to complicated. Therefore consider the series expansion of $(1-x)^n$ which is due the Binomial Theorem given by

$$sum_{k=0}^nbinom nk (-x)^k(1)^{n-k}=sum_{k=0}^nbinom nk (-x)^k$$

Further notice that the binomial coefficient can be written as

$$binom nk = frac{n!}{k!(n-k)!}$$

Putting these two together yields to

$$sum_{k=0}^nfrac{n!}{k!(n-k)!}(-x)^k=n!sum_{k=0}^nfrac{(-x)^k}{k!(n-k)!}$$

For $x=1$ this equals your given formula. Now look at the original term we have expanded. Can you finish it from hereon?

$sum_{k=0}^n frac{(-1)^{-k}}{k!(n-k)!}=1/(n!)sum_{k=0}^n frac{n!(-1)^{k}}{k!(n-k)!}=frac{1}{n!}sum_{k=0}^n(-1)^{k}{nchoose k}=frac{1}{n!}(1-1)^n=0.$