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

1 Answer
Feb 28, 2017

Answer:

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

Explanation:

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#