# Math Help - Sigma notation and sums

1. ## Sigma notation and sums

I have som question about sums never done them before, didn´t know where to post it its a analysis course so they cant be to misplaced (i hope). Am I on the right track or doing someting completely wrong? All tips appreciated.. $\Rightarrow$

Write the following using Sigma notation:
$1*2*3+2*3*4+3*4*5+...+101*102*103$

$\sum_{k=1}^{101}(k)(k+1)(k+2)$
2)
$1*2+2*2^2+3*2^3+...+10*2^{10}$

$\sum_{k=1}^{10}k\times2^k$
3)
then we have $1+4+9+16...+100$
$\sum_{k=1}^{10}(n^k)$
and finaly
4)
$\frac{1}{1\times 3}+\frac{1}{2\times 4}+\frac{1}{3\times 5}...$

$\sum_{k=1}^{\infty} \frac{1}{k\times (k+2)}$

Actually what does this mean
$\sum_{k=1}^{n} \frac{k}{n+k}$ n is suppose to be the final value. The exercis said write without Sigma notation. My question is does n=k so its $\frac{1}{1+1}+\frac{2}{2+2}+\frac{3}{3+3}$ ?

2. Originally Posted by Henryt999
I have som question about sums never done them before, didn´t know where to post it its a analysis course so they cant be to misplaced (i hope). Am I on the right track or doing someting completely wrong? All tips appreciated.. $\Rightarrow$

Write the following using Sigma notation:
$1*2*3+2*3*4+3*4*5+...+101*102*103$

$\sum_{k=1}^{101}(k)(k+1)(k+2)$
Good.

2)
$1*2+2*2^2+3*2^3+...+10*2^{10}$

$\sum_{k=1}^{10}k\times2^k$
Good

3)
then we have $1+4+9+16...+100$
$\sum_{k=1}^{10}(n^k)$
No. $4= 2^2$, [tex]9= 32, $16= 4^2$, ..., $100= 10^2$
The exponent is a constant, 2, not the index. And you don't say what "n" is!

and finaly
4)
$\frac{1}{1\times 3}+\frac{1}{2\times 4}+\frac{1}{3\times 5}...$

$\sum_{k=1}^{\infty} \frac{1}{k\times (k+2)}$
Good! You only need to redo (2).

Actually what does this mean
$\sum_{k=1}^{n} \frac{k}{n+k}$ n is suppose to be the final value. The exercis said write without Sigma notation. My question is does n=k so its $\frac{1}{1+1}+\frac{2}{2+2}+\frac{3}{3+3}$ ?
No. n is a fixed number and k runs from 1 to n. This is
$\frac{1}{n+1}+ \frac{2}{n+2}+ \cdot\cdot\cdot+ \frac{n-1}{n+ n-1}+ \frac{n}{n+ n}$.