Question

Find in terms of the positive integer k, the sum of the integers from 2k to 4k inclusively

Answer 1

just use the formula (twice)

\[\sum_{x=1}^{n}x=\frac{n(n+1)}{2}\]

Answer 2

how did you know that is the formula to use?

Answer 3

think of it like this

Answer 4

how do you know the conjecture is 'i'?

Answer 5

Whenever there is a question about adding up consecutive integers (like 1+2+3+4+...+15, or 22+23+24+...+50) that formula should be the first thing that pops in your mind.

Answer 6

the 'i' is just a dummy variable, a place holder. If you look at zarkon's formula, he used x instead. It means the same thing. we could pick any letter we want for that purpose :)

Answer 7

but the 'i ' beside the summation symbol what does it mean?

Answer 8

\[\sum_{\Omega=1}^{\Gamma}\Omega=\frac{\Gamma(\Gamma+1)}{2}\]

;)

Answer 9

@ zarkon lol that makes it look awesome
The 'i' next to the summation is the formula (or function) you are plugging each value of the index into. For example:
\[\sum_{i=1}^{5}i = 1+2+3+4+5\]
or
\[\sum_{i=5}^{10}i^2 = 5^2+6^2+7^2+8^2+9^2+10^2\]
In general:
\[\sum_{i=a}^{b}f(i) = f(a)+f(a+1)+f(a+2)+\cdots +f(b-1)+f(b)\]

Answer 10

i hope thats not too confusing >.< im sry if it is.

Answer 11

how do you know that is the formula to use? it only tells me from 2k to 4k which really doesn't tell me much..

Answer 12

Well, you are adding up all the integers in between 2k and 4k, so in my mind im thinking:
\[2k+(2k+1)+(2k+2)+(2k+3)+\cdots+(4k-2)+(4k-1)+4k\]

Answer 13

if k was equal to say...3, then we would be adding the numbers from 6 to 12:
\[6+7+8+9+10+11+12\]

Answer 14

so i can say it's like an adding up series and that is why we use that formula (1/2)(n)(n+1) which is like an AP summation?

Answer 15

right right :)

Answer 16

hmmm i think i get it.. i shall try it out =)

Answer 17

joemath! now the question asks for the sum of all odd integers lying between 2k and 4k what do i do???

Answer 18

hmm...we need a formula for the sum of the odd integers...unfortunately i dont know one off the top of my head, maybe Zarkon does? i'll have to work out one on paper. one sec.

Answer 19

\[3k^2\]

Answer 20

Alright i came up with a formula for the sum of all the odd numbers up to a given odd number (say, n). its:
\[1+3+5+7+\cdots+(n-2)+n = \frac{(n+1)^2}{4}\]
Note that n must be odd.
For example:
\[1+3+5+7 = \frac{(7+1)^2}{4} = \frac{8^2}{4} = \frac{64}{4} = 16\]

Answer 21

\[\sum_{i=k}^{2k-1}{2i+1}\]

first term is 2k+1 and last term is 2(2k-1)+1=4k-2+1=4k-1. and it is counting the odd numbers

Answer 22

thing is we must have a fromula for summing up odd numbers?

Answer 23

\[\sum_{i=k}^{2k-1}{2i+1}=3k^2\]

Answer 24

ah, that what we need, nice Zarkon.

Answer 25

sorry it took so long...I got distracted by the dead bird my wife found in the back yard :(

Answer 26

first odd term is 2k+1?

Answer 27

>.<

Answer 28

yes, 2k is even (because its divisible by 2). So 2k+1 has to be the first odd term.

Answer 29

ok... so we have to sum up from 2k to 4k odd terms... first term is 2k+1... then how do i proceed??

Answer 30

That summation zarkon posted works out perfectly:
\[\sum_{i=k}^{2k-1}2i+1 = (2k+1)+(2(k+1)+1)+(2(k+2)+1)+\cdots + (2(2k-1)+1)\]
=\[(2k+1)+(2k+3)+(2k+5)+\cdots +(4k-1)\]
It hits all the odd numbers.
So we just have to figure out the sum, which i'll do on paper and post.

Answer 31

sigh.. i don't understand..

Answer 32

can you be more specific about what you don't understand?