Intereting Posts

On the equation $3x^3 + 4y^3 + 5z^3 = 0$
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Using induction to prove an equality in harmonic numbers
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Let $a,b \in \mathbb{Q^{+}}$ then $\sqrt{a} + \sqrt{b} \in \mathbb{Q} $ iff $\sqrt{a} \in \mathbb{Q}$ and $\sqrt{b} \in \mathbb{Q}$
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Prove that two consecutive square free numbers can have arbitrarily large gaps
Find all of the solutions of $z^4=2i$
Let $W$ be a subspace of a vector space $V$ . Show that the following are equivalent.

Show that $$\sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0$$

So for odd $n$ we have an even number of terms. So $\binom{n}{k} = \binom{n}{n-k}$ which have opposite signs. Thus the sum is $0$.

For even $n$ we have that $$\sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = \binom{n}{0}+\sum_{k=1}^{n-1} (-1)^{k} \binom{n}{k} + \binom{n}{n}$$

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Now $$\sum_{k=1}^{n-1} (-1)^{k} \binom{n}{k} = \sum_{k=1}^{n-1} (-1)^{k} \left[\binom{n-1}{k} + \binom{n-1}{k-1} \right]$$

What would that sum be in the square brackets?

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- Proving $\sum_{k=1}^n k\cdot k! = (n+1)!-1$ without using mathematical Induction.
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Here is an alternate way:

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

by the binomial theorem.

Another way for combinatorially-minded people:

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

is the number of ways to flip n coins and get an even number of heads, minus the number of ways to flip n coins and get an odd number of heads. Since the parity of the number of heads will always come down to the last coin flipped, and heads/tails are of course equally likely at that point, the sum evaluates to 0.

You should know that

$$

(a+b)^n=\sum_{k=0}^n \binom nk a^kb^{n-k}.

$$

When $b=1$, this says

$$

(1+a)^n=\sum_{k=0}^n \binom nk a^k.

$$

So now you just need to consider the case where $a=-1$.

For odd $n$, the OP gave a **bijective** proof of the result. The following is a bijecive proof that works for all $n$.

Let $A=\{1,2,\dots, n\}$. Let $\mathcal{E}$ consist of the subsets of $A$ of even cardinality, and let $\mathcal{O}$ consist of the subsets of $A$ of odd cardinality. We produce an explicit bijection $\varphi: \mathcal{E} \to \mathcal{O}$.

If $S\in \mathcal{E}$ and $1\notin S$, let $\varphi(S)=S\cup\{1\}$.

If $S\in \mathcal{E}$ and $1\in S$, let $\varphi(S)=S\setminus\{1\}$.

**Comment:** For the sake of symmetry, it is probably better to define $\varphi(S)$ as above, but for *any* subset $S$ of $A$. Then $\varphi$ is an *involution* on $P(A)$ that interchanges sets of even cardinality with sets of odd cardinality.

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

We have, since $n-1$ is odd

\begin{align*}\sum_{k=1}^{n-1}(-1)^k\left[\binom{n-1}k+\binom{n-1}{k-1}\right]&=\sum_{k=1}^{n-1}(-1)^k\binom{n-1}k-\sum_{j=0}^{n-2}(-1)^j\binom{n-1}j\\

&=-1+\binom{n-1}{n-1}(-1)^{n-1}\\

&=-2,

\end{align*}

hence $$\sum_{k=0}^n\binom nk(-1)^k=\binom n0 -2+\binom nn =0.$$

One can also give a topological proof of the binomial identity $\sum_{i=0} ^{n} (-1)^i \binom{n}{i}=0$. Consider the standard $(n-1)$-simplex $\Delta^{n-1}$. This has $\binom{n+1}{i+1}$ $i$-simplicies, hence the Euler characteristic of $\Delta^{n-1}$ is

$$\chi(\Delta^{n-1})=\sum_{i=0} ^{n-1} (-1)^i \binom{n}{i+1}.$$

The Euler characteristic is homotopy invariant and $\Delta^{n-1}\simeq *$, so we must have

$$\sum_{i=0} ^{n-1} (-1)^i \binom{n}{i+1}=1,$$

or

$$\sum_{i=0} ^{n-1} (-1)^i \binom{n}{i+1}-\binom{n}{0}=0$$

since $\binom{n}{0}=1$.

Multiplying both sides by $-1$ yields the identity

$$\sum_{i=0} ^{n} (-1)^i \binom{n}{i}=0.$$

Another approach:

For the difference operator $\Delta f (m) = f(m+1)-f(m)$, the $n$th iteration is

$$\Delta^n f(m)=\sum_{k}(-1)^{n-k}\binom nk f(m+k)$$

Clearly, your sum is the $n$th difference of the constant function $f\equiv 1$, so of course it is zero for $n\ge 1$ because already $f(m+1)-f(m) =0 $.

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