1. ## Evaulating Riemann Sums

I have another questions about thes,e no very good with them as you may seen from my previous posts . Anyhow, here is the problem below.

$

\sum _{i=1}^n(ni + \sum _{j=i}^nnj)
$

Ok, so what i do know is that i should expand these summations and then look for a pattern? Then, i assume that im suppose to derive sometype of formula? Im kind of unsure whats the whole purpose of riemann sums ecept for just summing a whole bunch of numbers.

Kind of like nested integrals i was told that you should evaluate the inner summation, and then work your way to the outer.

So when i expand the inner loop i get $n+2n+3n+4n +...+ jn, and\ this\ is\ when\ j=n$

2. Expand outer summation
$n+2n+3n+4n+ ... +ni, and\ this\ is\ when\ i = n$

Im confused on what to do next.

PS: I will have one Riemann sum problem to ask tonight, but i will make sure to try and get far on it as much as possible before i post to the forums. Thanks!

2. Originally Posted by tokio
I have another questions about thes,e no very good with them as you may seen from my previous posts . Anyhow, here is the problem below.

$

\sum _{i=1}^n(ni + \sum _{j=i}^nnj)
$
n is fixed and can be take outside the sums:
$n\sum_{i=1}^n(i+ \sum_{j=1}^n j)$

Now, a very useful formula, I am surprised you didn't already know it:
$\sum_{j=1}^n j= \frac{n(n+1)}{2}$

You can prove that easily by writing the sum as 1+ 2+ 3+ ...+ (n-2)+ n-1+ n, reversing it as n+ (n-1)+ (n-2) +...+ 3+ 2+ 1 and adding each term to the one above: n+ 1, n-1+ 2= n+ 1, n-2+ 3= n+1, etc. Each pair adds to n+1 and there are n terms- that sum is n(n+1). Since we added the sum twice, divide by 2 to get the formula above.

So now you have $\sum_{i=1}(i+ \frac{n(n+1)}{2})$.
Break that into two sums: $\sum_{i=1}^n i$ is, of course, $\frac{n(n+1)}{2}$ again. And $\sum_{i= 1}^n \frac{n(n+1)}{n}$ is just a constant added to itself n times- that is $n(\frac{n(n+1)}{2})= \frac{n^2(n+1)}{2}$

Ok, so what i do know is that i should expand these summations and then look for a pattern? Then, i assume that im suppose to derive sometype of formula? Im kind of unsure whats the whole purpose of riemann sums ecept for just summing a whole bunch of numbers.

Kind of like nested integrals i was told that you should evaluate the inner summation, and then work your way to the outer.

So when i expand the inner loop i get $n+2n+3n+4n +...+ jn, and\ this\ is\ when\ j=n$

2. Expand outer summation
$n+2n+3n+4n+ ... +ni, and\ this\ is\ when\ i = n$

Im confused on what to do next.

PS: I will have one Riemann sum problem to ask tonight, but i will make sure to try and get far on it as much as possible before i post to the forums. Thanks!