When does $1-2+3-4+5-\cdots +(-1)^{n+1}n$ reach $100$?

When does $1-2+3-4+5-\cdots +(-1)^{n+1}n$ reach $100$?

Question

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

1. $99$,

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1. $101$,

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1. $199$,

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1. $300$.

Suggestion

What is the sum of the terms of the sequence when $n=1$?

What about when $n=2$? $n=3$? $n=4$?

When will this sum reach $100$?

Or… what is the sum of the first $2n$ terms of the sequence?

Solution

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).