How do you find the partial sum of #Sigma 2n# from n=1 to 100?

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Answer 1 · 2018-02-20T00:14:19

# sum_(n=1)^100 \ 2n =10100 #

We seek the sum:

# S_100 = sum_(n=1)^100 \ 2n #

We can readily find the general sum:

# S_n = sum_(r=1)^n \ 2r #
# \ \ \ \ = 2 \ sum_(r=1)^n \ r #

And. we use the standard formula for the sum of the first #n# integers:

# sum_(r=1)^n \ r = 1/2n(n+1) #

Giving us:

# S_n = 2 * 1/2n(n+1) #
# \ \ \ \ = n(n+1) #

So with #n=100# we have:

# S_100 = 100(101) #
# \ \ \ \ \ \ \ = 10100 #