What is the smallest positive integer such that 1-2+3-4+5-6+...+(-1)^(n+1)n is more than or equal to 100?

Question

My attempt to answer this question was to split the sum of this sequence into the sum of other two sequences i.e. 1,3,5... and -2,-4,-6... which ended up giving me the wrong answer to the problem. I do not understand how this sequence ends up being the sum of 1,-1,2,-2, and so on. Could someone please explain how this sequence is the same as the sequence I just mentioned, and why splitting it into two sequences is wrong?

Answer 1

For even $n$ it's negative.

But for odd $n$ we obtain: $$-1-1-...-1+n\geq100$$ or $$-\frac{n-1}{2}+n\geq100$$ or $$n\geq199.$$

Answer 2

$1+(-2+3)+(-4+5)+(-6+7)+\dots= 1+1+1+1+\dots=100$. $100$ ones required. Last $99$ ones comes from two terms. Therefore number of terms required is $99\times2+1= 199$.