The smallest possible integer \(n\) such that \[1-2+3-4+5-6+\cdots +(-1)^{n+1}n \geq 100\] is

\(99\),

\(101\),

\(199\),

\(300\).

#### Approach 1

What’s the result of this sum after \(n\) steps? We get the sequence

\[1, -1, 2, -2, 3, -3, \cdots.\]

When do we first reach \(100\) in this sequence?

Clearly the \(199^{th}\) term will be the first one to give at least \(100\), so the answer is (c).

#### Approach 2

Looking at \((1-2)+(3-4)+(5-6)+\cdots\), we see that the sum of the first \(2n\) terms is \(-n\).

So adding the next term, we require \((2n+1) -n = 100\), which gives \(n = 99\).

Thus \(2n+1 = 199\), so the answer is (c).