# Thread: Another sums question =)

1. ## Another sums question =)

Some sums questions again:
Have some thoughts about if these solutions are correct..all comments highly appreciated
First one out is:
Write without Sigma notation
$\displaystyle \sum_{k=1}^{n}\frac{k}{n+k}$

well here is my shot should be:
$\displaystyle \frac{1}{n+1}+\frac{2}{n+2} +\frac{3}{n+3}$

Then we have a "calc the sum of stuff":
$\displaystyle 3+7+11+15+...+43$
well I figured since there is a 4 between each term why dont we: $\displaystyle (43+3)\times \frac{43-3}{4}$$\displaystyle = 460 We have another question: calculate the sum of:\displaystyle 1+3+5+...+(2n-1) well if I write it so the first number overlapswith the last 1+ .............3+ ...........5...........+(2n-1) (2n-1)+1 +(2(n-1))+3 + (2(n-2))+5 + (2n-1)+1 then \displaystyle \frac{2n\times n}{2} = \frac{2n^2}{2} = n^2 Then a more difficult one calculate the sum of: \displaystyle a-ak+ak^2-ak^3+...+a(-k)^n \displaystyle a(1-k(1-k)-k^3(1-k)-k^{n-1}(1-k)) hey lets divide by a \displaystyle a(1-k)(1-k-k^3-k^5-k^{n-1}) oohh look divide by (1-k) \displaystyle (a-ak)(1-k-k^3-k^5-k^{n-1}-k^n) hej lets add \displaystyle (a-ak)(1-(n^2-k)) = a- an^2-2ak+ak^2+akn^2 turned out like this. 2. Originally Posted by Henryt999 Some sums questions again: Have some thoughts about if these solutions are correct..all comments highly appreciated First one out is: Write without Sigma notation \displaystyle \sum_{k=1}^{n}\frac{k}{n+k} well here is my shot should be: \displaystyle \frac{1}{n+1}+\frac{2}{n+2} +\frac{3}{n+3} Don't understand this one: you wrote the first three summands of the above series...so? Then we have a "calc the sum of stuff": \displaystyle 3+7+11+15+...+43 well I figured since there is a 4 between each term why dont we: \displaystyle (43+3)\times \frac{43-3}{4}$$\displaystyle = 460$

This is the sum of an arithmetic series with first element 3 and common difference 4. If $\displaystyle a_1, a_2,\ldots,a_n$ is an arithmetic series, with common difference $\displaystyle d=a_n-a_{n-1}\,\,\forall n$ , then the sum of its first consecutive n terms is $\displaystyle S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$ .

We have another question: calculate the sum of:$\displaystyle 1+3+5+...+(2n-1)$
well if I write it so the first number overlapswith the last
1+ .............3+ ...........5...........+(2n-1)
(2n-1)+1 +(2(n-1))+3 + (2(n-2))+5 + (2n-1)+1
then $\displaystyle \frac{2n\times n}{2} = \frac{2n^2}{2} = n^2$

Correct. Again, this is the sum of an arithmetic series with difference 2 and first element 1

Then a more difficult one calculate the sum of:
$\displaystyle a-ak+ak^2-ak^3+...+a(-k)^n$

$\displaystyle a(1-k(1-k)-k^3(1-k)-k^{n-1}(1-k))$ hey lets divide by a
$\displaystyle a(1-k)(1-k-k^3-k^5-k^{n-1})$ oohh look divide by (1-k)
$\displaystyle (a-ak)(1-k-k^3-k^5-k^{n-1}-k^n)$ hej lets add
$\displaystyle (a-ak)(1-(n^2-k)) = a- an^2-2ak+ak^2+akn^2$ turned out like this.

The sume of the first n consecutive elements of a geometric series $\displaystyle a_1,a_2,\ldots,a_n$ with common quotient $\displaystyle q=\frac{a_n}{a_{n-1}}\,\,\forall n$ is given by $\displaystyle S_n=a_1\frac{q^n-1}{q-1}$

Here you have $\displaystyle a_1=1\,,\,\,q=-k$ so now just substitute.

Tonio