What is the sum of #sum_(r=1)^(100)(2r+1)#?

Question

I know I could just put values in and add them up, but what is a more elegant why of doing this. It seems impossibly difficult, or at least it does to me.

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Answer 1 · 2018-02-01T17:58:19

#10200#

#sum_(r=1)^(100)(2r+1)#

First we need to know two of the most usful properties of sigma notation.

#sum_(r=1)^(n)[(f(r)+g(r)]=sum_(r=1)^(n)f(r)+sum_(r=1)^(n)g(r)#

#sum_(r=1)^(n)af(r)=asum_(r=1)^(n)f(r)#

We know the sum of the first #n# integers is given by:

#1/2n(n+1)#

So:

#sum_(r=1)^(100)(2r+1)=sum_(r=1)^(100)2r+sum_(r=1)^(100)1#

#sum_(r=1)^(100)2r=2sum_(r=1)^(100)r#

This is the sum of the first 100 integers multiplied by 2:

#:.#

#2sum_(r=1)^(100)r=2*1/2*100(101)#

#sum_(r=1)^(100)1#

This is just 100:

#sum_(r=1)^(100)1=100#

i.e. #sum_(r=1)^(n)(a)=nacolor(white)(88)# where #a# is a constant.

Putting everything together:

#2*1/2*100(101)+100=#

#10100+100=10200#

#sum_(r=1)^(100)(2r+1)=10200#