Find closed form for $2 \times (n + (n + 1) + \cdots + (2n - 1))$

Question

How can we find a closed form for this sum: $$2 \times (n + (n + 1) + \cdots + (2n - 1))$$?

Example: $$(7+6+5+4)+(4+5+6+7)=\frac{3}{4} 8^2 - \frac{1}{2} 8$$

Answer 1

As I understand it, you want a formula to obtain twice the sum of the numbers $n + (n + 1) + \cdots + (2n - 1)$. Let's call that $S$.

First observe that the sum of the natural numbers from $1$ to $m$ (inclusive) is $$\frac {m(m+1)} 2$$

You can think of your example as follows: $$\underbrace{(7 + 6 + 5 + 4 + 3 + 2 + 1)}_{\displaystyle\frac {7 \cdot 8} 2} - \underbrace{(3 + 2 + 1)}_{\displaystyle\frac {3 \cdot 4} 2}$$

Let $k$ be the upper bound of your sum; the lower one is $k/2$. In your example $k$ is $8$.

Then: $$S = 2 \times \left(\frac {k(k-1)} 2 - \frac {\frac k2\left(\frac k2 - 1\right)} 2\right) = k^2 - k - \frac {k^2} 4 + \frac k2 = \boxed{\frac 34 k^2 - \frac 12 k}\qquad\square$$

Answer 2

The sum

$$\sum_{k = n}^{2n - 1} k = n + (n + 1) + \cdots + (2n - 1)$$

is a partial sum of an arithmetic series with initial term $a_1 = n$, $n$th term $a_n = 2n - 1$, and common difference $1$. The $n$th partial sum of an arithmetic series with initial term $a_1$, $n$th term $a_n$, and common difference $d$ is

$$s_n = \frac{n(a_1 + a_n)}{2}$$

Substituting $n$ for $a_1$ and $2n - 1$ for $a_n$ yields

$$s_n = \frac{n(n + 2n - 1)}{2} = \frac{n(3n - 1)}{2} = \frac{3n^2 - n}{2}$$

Since you want to find $2[n + (n + 1) + \cdots + (2n - 1)]$, we double this result to obtain

$$2[n + (n + 1) + \cdots + (2n - 1)] = 2\sum_{k = n}^{2n - 1} k = 2s_n = 3n^2 - n$$