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.

848 views around the world

You can reuse this answer
Creative Commons License

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#