# How do you find the partial sum of Sigma (2n-1) from n=1 to 400?

Feb 28, 2017

${\sum}_{n = 1}^{400} \left(2 n - 1\right) = 16000$

#### Explanation:

We have that using the distributive property of the sum:

${\sum}_{n = 1}^{N} \left(2 n - 1\right) = 2 {\sum}_{n = 1}^{N} n - {\sum}_{n = 1}^{N} 1$

The sum of the first $N$ integers is given by Gauss' formula:

${\sum}_{n = 1}^{N} n = \frac{N \left(N + 1\right)}{2}$

while the sum of $N$ times the unity is clearly $N$:

${\sum}_{n = 1}^{N} 1 = N$

So:

${\sum}_{n = 1}^{N} \left(2 n - 1\right) = 2 \frac{N \left(N + 1\right)}{2} - N = {N}^{2} + N - N = {N}^{2}$

For $N = 400$:

${\sum}_{n = 1}^{400} \left(2 n - 1\right) = {400}^{2} = 16000$