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

Question

8426 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-28T16:27:59

$sum_(n=1)^400 (2n-1) = 16000$

We have that using the distributive property of the sum:

$sum_(n=1)^N (2n-1) = 2 sum_(n=1)^N n -sum_(n=1)^N 1$

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

$sum_(n=1)^N n = (N(N+1))/2$

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

$sum_(n=1)^N 1 = N$

So:

$sum_(n=1)^N (2n-1) = 2(N(N+1))/2 -N = N^2+N-N = N^2$

For $N=400$ :

$sum_(n=1)^400 (2n-1) =400^2 = 16000$

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