How to compute $\sum_{k =1}^{100}(-1)^k$ Today I tried to compute $$ \sum_{k =1}^{100}(-1)^k $$
Is there a way to find the result more quickly ? 
Below if my attempt to find the result. 
Especially without considering the case of odd and even numbers like I did ?

Let's consider the following sum: 
$$ \sum_{k=1}^{n}(-1)^k = (-1)^1 + (-1)^2 + (-1)^3 + ...  +(-1)^n $$
If $n$ is even then $\frac n2 $ terms have an even exponent and $\frac n2 $ terms have an odd exponent. 
$$(-1)^p = -1   \tag{when $p$ is odd}   $$
$$ (-1)^p = 1  \tag{when $p$ is even}
$$
Then when $n$ is even
$$\begin{align}
  \sum_{k=1}^{n}(-1)^k &= \frac n2 \times(-1) +   \frac n2 \times 1  \\
  & =  \frac n2  - \frac n2 \\  
  & =   0 \\ 
\end{align}$$
100 is an even number, so 
$$ \sum_{k =1}^{100}(-1)^k = 0 $$
 A: Note that $(-1)^{2k-1}=-1$ and $(-1)^{2k}=1$ for all $k\geq1$. Hence the sum upto any even power should equal $0$.
A: The easiest way to find the result is to look at the first few partial sums
$$ -1, 0, -1, \ldots $$
and identify the pattern. Everything beyond that is just seeking to give a more rigorous justification of it.
A: Since $(-1)^{2n}=1$ and $(-1)^{2n-1}=-1$, we have
$$\sum_{k=1}^n (-1)^k = -1+1-1+1-\cdots \pm 1=\left\{\begin{array}{ll}-1 & \text{if}\,n\,\text{is odd}\\
0 & \text{if}\,n\,\text{is even}\end{array}\right.$$
A: $$\sum_{k=1}^{n}(-1)^k=\sum_{k=1}^{n}(-1)(-1)^{k-1}=-\sum_{k=1}^{n}(-1)^{k-1}=$$
$$=-\sum_{k=0}^{n-1}(-1)^{k}=-\frac{1-(-1)^{n}}{1-(-1)}=\frac{(-1)^{n}-1}{2}$$
for $n=100$
$$\sum_{k=1}^{100}(-1)^k=\frac{(-1)^{100}-1}{2}=\frac{1-1}{2}=0$$
A: yes the series has a direct summation formula: 
$$
S_k=\sum_{i=1}^k (-1)^i={1 \over 2}\Big((-1)^k-1\Big)
$$
you can prove this easily using induction.
$$\begin{aligned}
\mbox{k=1: }\ \ \ &true\\
\mbox{k }\to\mbox{ k+1: }\ \ \ &S_{k+1}=S_{k}+(-1)^{k+1}&={1 \over 2}\Big((-1)^k-1\Big)+(-1)^{k+1} =\\
& &={1 \over 2}\Big((-1)^k+2(-1)^{k+1}-1\Big) =\\
& &={1 \over 2}\Big((-1)^k(1+2(-1)^{1})-1\Big) =\\
& &={1 \over 2}\Big((-1)^k(1-2)-1\Big) =\\
& &={1 \over 2}\Big((-1)^k(-1)-1\Big) =\\
& &={1 \over 2}\Big((-1)^{k+1}-1\Big) \ \ \ \square
\end{aligned}
$$
