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

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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#