Question

evaluate the series (1 + (2k/n))(1/n) from k=1 to n

Answer 1

\[\sum_{k=1}^n(1+\frac{2k}n)(\frac1n)\]?

Answer 2

yup

Answer 3

n is a constant so let's treat it as such, meaning we can pull it out of the series when multiplying\[\sum_{k=1}^n(1+\frac{2k}n)(\frac1n)=\frac1{n^2}\sum_{k=1}^n(n+2k)\]so far so good?

Answer 4

yup

Answer 5

now any idea how we go about handling what's left over?
remember that n is still a constant, while k will be changing...

Answer 6

we never learnt this in the first place so ....

Answer 7

consider the series\[\sum_{k=1}^n5\]what is the sum of that?

Answer 8

5?

Answer 9

no, how about\[\sum_{k=1}^35\]think about what the sum sign means

Answer 10

\[\sum_{k=1}^35=5+5+5=3(5)\]

Answer 11

15?

Answer 12

right :)

Answer 13

so the sigma means to do something an n number of times?

Answer 14

it means to add up n number of those terms, so yeah
the specific "something" that you are doing n times is adding the terms together

Answer 15

so notice what happened there\[\sum_{k=1}^35=5+5+5=3(5)=15\]so what about in general?\[\sum_{k=1}^n5\]since 5 will occur n times, the sum will be...?

Answer 16

5*n?

Answer 17

yes :)

Answer 18

now, remembering that n is a constant, what does that mean about\[\sum_{k=1}^nn=?\]

Answer 19

n^2?

Answer 20

yes :) very good
so now let's return to our problem...

Answer 21

\[\sum_{k=1}^n(1+\frac{2k}n)(\frac1n)=\frac1{n^2}\sum_{k=1}^n(n+2k)\]we can split up the next part into two remaining sums:\[\frac1{n^2}\left(\sum_{k=1}^nn+\sum_{k=1}^n2k\right)=\frac1{n^2}\left(n^2+\sum_{k=1}^n2k\right)=1+\frac1{n^2}\sum_{k=1}^n2k\]still with me?

Answer 22

yup

Answer 23

now what can we do about\[\sum_{k=1}^n2k\]?
first notice that 2 is a constant, so what can we do with it?

Answer 24

take it out

Answer 25

right, so we have\[\frac1{n^2}\left(\sum_{k=1}^nn+\sum_{k=1}^n2k\right)=\frac1{n^2}\left(n^2+\sum_{k=1}^n2k\right)=1+\frac2{n^2}\sum_{k=1}^nk\]so now it all comes down to\[\sum_{k=1}^nk=1+2+3+...+(n-1)+n\]for which there is a very well-known formula (attributed to the mathematician Gauss) that says the the sum of the first n natural numbers is\[{n(n+1)\over2}\]and that is precisely what we have with our sum so use that formula for the last sum and you're ready to do the final algebra needed to solve it

Answer 26

okay, i see

Answer 27

you should really try to prove that theorem if/when you learn mathematical induction. It's quite useful.

Answer 28

so the answer would be 1 + (n+1)/n ?

Answer 29

yep :)
nice job!

Answer 30

hmm thats funny, our teacher got 1 + ((n+2)(n))/2

Answer 31

that is funny, but instead of laughing I want a second opinion to see who's actually wrong

@amistre64 am I wrong about this series?

Answer 32

i was confused about where she got the +2 from. i get the logic behind yours though

Answer 33

@asnaseer did I mess up?

Answer 34

@Callisto

Answer 35

the answer [1 + (n+1)/n = 1 + 1 + 1/n = 2 + 1/n] is correct

Answer 36

oh good, so I guess either you copied part of the problem wrong, or the teacher was wrong @halcyon98

Answer 37

and your approach is also correct

Answer 38

thanks asnaseer :)

Answer 39

yw :)

Answer 40

i think she was wrong because here answers were way off on a bunch. thanks alot for clearing that up

Answer 41

I just entered this in wolfram as a double check: http://www.wolframalpha.com/input/?i=sum+%281+%2B+%282k%2Fn%29%29%281%2Fn%29+k%3D1+to+n

answer still correct :)

Answer 42

great, thanks!

Answer 43

yw :)