A Simple Pattern

$\dfrac{n(n+1)}{2} > 1001$ $n > 44.2$ $\boxed{n = 45}$

So we start by looking at the pattern. We see that the 1st term is the last 1, the 3rd term is the last 2, the 6th term is the last 3, and so on. So we can guess that the last term for the number n is 1+2+3+...(n-1)+n.

Solution 1: Brute Force

We now start by testing some values of n. For 10, the answer is 55.

For 20, the answer is 210

For 30, the answer is 465

For 40, the answer is 840

For 50, the answer is 1275

Aha! So the answer has to be between 40 and 50!

Now we test 45, and we get 1035

So the last 45 in the sequence is the 1035th term

We now test 44, and we get 990

So the last 44 in the sequence is the 990th term

So, since the 1001th term is after the 990th term and before the 1035th term, the answer is 45.

Solution 2: inequalities

We can also have solved this without trial and error.

1+2+3+...(n-1)+n>1001

n(n-1)/2>1001

n>44.2

n=45 (Because it has to be the smallest value that satisfies the inequality)

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