Question

The sum of the first n consecutive natural numbers is (n(n+1). For example, 1+2+3+4=10=(4(4+1))/2. Find all values of n with the property that this sum lies between 150 and 200.

Answer 1

You mean \(\dfrac{1}{2}n(n+1)\).

So you have to solve \(150 < \dfrac{1}{2}n(n+1)<200\).

I would try to solve these two equations:

1). \(\dfrac{1}{2}n(n+1)=150\)
2). \(\dfrac{1}{2}n(n+1) =200\).

If n comes out as a decimal number, round it off appropriately.

Answer 2

(n(n+1))/2*

Answer 3

If you begin to solve the equations, you get:

1). \(n(n+1)=300 \Leftrightarrow n^2+n-300=0\)
2). \(n(n+1)=400 \Leftrightarrow n^2+n-400=0\)

You can use the Quadratic Formula to solve the equations.
Each has two solutions, one positive and one negative.
You can ignore the negative ones, because we are talking about positive numbers.