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).
