Welcome to our community

Be a part of something great, join today!

Ally Samaniego's question at Yahoo! Answers regarding partial sums

  • Thread starter
  • Admin
  • #1


Staff member
Feb 24, 2012
Here are the questions:

How do I find the nth sum of these functions (calculus I)?

My professor explained it very poorly and I have no idea how to solve these.
If possible, could you walk me through it and give me a good way to think about these to solve them properly? It's greatly appreciated, my exam is next week.

1) Sum of i=1 to n of (4-9i)

2) Sum of i=1 to n of (i+1)(i+4)

3) Sum of i=1 to n of (i^3 - i - 2)

4) Sum of i=1 to n of 3/n(i/n)^2

I don't need all of them done, I just need someone to explain how to do these. I'm so confused. I really appreciate your help!
I have posted a link there to this thread so the OP can see my work.
  • Thread starter
  • Admin
  • #2


Staff member
Feb 24, 2012
Hello Ally Samaniego,

To work these problems, you will need the following formulas:

(1) \(\displaystyle \sum_{k=1}^n(1)=n\)

(2) \(\displaystyle \sum_{k=1}^n(k)=\frac{n(n+1)}{2}\)

(3) \(\displaystyle \sum_{k=1}^n\left(k^2 \right)=\frac{n(n+1)(2n+1)}{6}\)

(4) \(\displaystyle \sum_{k=1}^n\left(k^3 \right)=\frac{n^2(n+1)^2}{4}\)

You will also need:

(5) \(\displaystyle \sum_{k=a}^b\left(c\cdot f(k) \right)=c\cdot\sum_{k=a}^b\left(f(k) \right)\) where $c$ is an arbitrary constant.

(6) \(\displaystyle \sum_{k=a}^b\left(f(k)\pm g(k) \right)=\sum_{k=a}^b\left(f(k) \right)\pm\sum_{k=a}^b\left(g(k) \right)\)

Now, let's look at the problems:

1.) \(\displaystyle S_n=\sum_{i=1}^n(4-9i)\)

Using (6) and (5), we may write:

\(\displaystyle S_n=4\sum_{i=1}^n(1)-9\sum_{i=1}^n(i)\)

Using (1) and (2), we may write:

\(\displaystyle S_n=4n-9\frac{n(n+1)}{2}\)

Combining terms, we obtain:

\(\displaystyle S_n=\frac{8n-9n(n+1)}{2}=\frac{n(8-9(n+1))}{2}=-\frac{n(9n+1)}{2}\)

2.) \(\displaystyle S_n=\sum_{i=1}^n\left((i+1)(i+4) \right)\)

Expanding the summand, we obtain:

\(\displaystyle S_n=\sum_{i=1}^n\left(i^2+5i+4 \right)\)

Using (6) and (5), we may write:

\(\displaystyle S_n=\sum_{i=1}^n\left(i^2 \right)+5\sum_{i=1}^n(i)+4\sum_{i=1}^n(1)\)

Applying (1), (2) and (3), we obtain:

\(\displaystyle S_n=\frac{n(n+1)(2n+1)}{6}+5\frac{n(n+1)}{2}+4n\)

Combining terms, we find:

\(\displaystyle S_n=\frac{n(n+1)(2n+1)+15n(n+1)+24n}{6}= \frac{n\left((n+1)(2n+1)+15(n+1)+24 \right)}{6}= \frac{n\left(2n^2+3n+1+15n+15+24 \right)}{6}= \frac{2n\left(n^2+9n+20 \right)}{6}= \frac{n(n+4)(n+5)}{3}\)

3.) \(\displaystyle S_n=\sum_{i=1}^n\left(i^3-i-2 \right)\)

Using (6) and (5), we may write:

\(\displaystyle S_n=\sum_{i=1}^n\left(i^3 \right)-\sum_{i=1}^n(i)-2\sum_{i=1}^n(1)\)

Using (1), (2) and (4), we have:

\(\displaystyle S_n=\frac{n^2(n+1)^2}{4}-\frac{n(n+1)}{2}-2n\)

Combining terms, we obtain:

\(\displaystyle S_n=\frac{n^2(n+1)^2-2n(n+1)-8n}{4}=\frac{n\left(n(n+1)^2-2(n+1)-8 \right)}{4}=\frac{n\left(n^3+2n^2+n-2n-2-8 \right)}{4}=\frac{n\left(n^3+2n^2-n-10 \right)}{4}\)

4.) \(\displaystyle S_n=\sum_{i=1}^n\left(\frac{3}{n}\left(\frac{i}{n} \right)^2 \right)\)

Using (5), we may write:

\(\displaystyle S_n=\frac{3}{n^3}\sum_{i=1}^n\left(i^2 \right)\)

Using (3), we obtain:

\(\displaystyle S_n=\frac{3}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{2n^2}\)