How do I simplify sigma notation to find the sum in terms of n only?

[Youngster]

Homework Statement

I'm actually asked to calculate the area under the curve 5x + x² over the interval [0,1] using Riemann Sums. I found the formula for the Riemann sum over the interval, it being the following:

\sum^{n}_{k=1} (\frac{5k}{n}+\frac{k^{2}}{n^{2}})(\frac{1}{n})

However, I am asked to simplify the sigma notation to find the sum in terms of n only, which is where I'm currently stuck.

Homework Equations

None, I believe. Except perhaps the sum formulas for positive integers

The Attempt at a Solution

I actually just went ahead and multiplied through and separated each term like so

\frac{5}{n^{2}}\sum^{n}_{k=1} k +\frac{1}{n^{3}}\sum^{n}_{k=1} k^{2}

I, however, was left with something completely different from the expected answers given. Am I going through this correctly?

 

[A. Bahat]

Yes, you are on the right track. Now use a closed-form expression for those two sums to get something only in terms of n; then you can take the limit as n goes to infinity.

 

[Youngster]

Okay, I figured out what I was doing wrong after going through my work - I neglected to multiply part of the numerator of the first sum by 3n, thus giving me something different. So after multiplying through, the end result should be:

\frac{15n^{3}+15n^{2}}{6n^{3}} + \frac{2n^{3}+3n^{2}+n}{6n^{3}}

This should simplify to:

\frac{17n^{3}+18n^{2}+n}{6n^{3}}

This could further be simplified by removing \frac{17n^{3}}{6n^{3}} to get

\frac{17}{6} + \frac{18n+1}{6n^{2}}, where an n is distributed out of the second term.

Then the limit should be \frac{17}{6} as n approaches infinity since the second term is equal to zero. :D

 

[Dick]

Okay, I figured out what I was doing wrong after going through my work - I neglected to multiply part of the numerator of the first sum by 3n, thus giving me something different. So after multiplying through, the end result should be:

\frac{15n^{3}+15n^{2}}{6n^{3}} + \frac{2n^{3}+3n^{2}+n}{6n^{3}}

This should simplify to:

\frac{17n^{3}+18n^{2}+n}{6n^{3}}

This could further be simplified by removing \frac{17n^{3}}{6n^{3}} to get

\frac{17}{6} + \frac{18n+1}{6n^{2}}, where an n is distributed out of the second term.

Then the limit should be \frac{17}{6} as n approaches infinity since the second term is equal to zero. :D

Sure, that's it.