Question

what is the summation formula for x^1?
if x^2,n(n+1)/6.

Answer 1

ahh variables lemme see if i can figure it out

Answer 2

nvm srry but ill give u a medal :)

Answer 3

they wont let me

Answer 4

this question could use some clarification
what is
\[x^1\] ? is it just
\[x\]? wand what does
if x^2, n(n+1)/6 mean?

Answer 5

summation formula for X power 2 is n(n+1)/6
so what I want to know is,whether the summation formula X power 1 same with X power 2 or not

Answer 6

ok first off i assume you mean
\[\sum_{k=1}^{\infty}k^2\] right? and if so, it is not
\[\frac{n(n+1)}{6}\] but rather it is
\[\sum_{k=1}^{\infty}k^2=\frac{n(n+1)(2n+1)}{6}\]

Answer 7

oh i see

Answer 8

then,its same with X power 1?

Answer 9

actually what i wrote was silly. i meant
\[\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\]

Answer 10

sum from 1 to n

Answer 11

for summing consecutive integers the formula is
\[\sum_{k=1}^n k =\frac{n(n+1)}{2}\]

Answer 12

how to solve this,use summation are of definition of area to find area under y=1+3x on(-1,5)

Answer 13

use which formula?

Answer 14

are you looking for the limit of the reimann sum?

Answer 15

area as the limit of sum

Answer 16

what a pain.

Answer 17

yeah,realy pain..

Answer 18

first of all we can find the answer using geometry.

Answer 19

the answer is 42 by plane geometry, so at least we know what we are looking for

Answer 20

now we can do this the smart way i suppose. we want to take the limit of the sum.
\[\lim_{n\rightarrow \infty} \sum_{k=1}^n f(x_k)\Delta x\]

Answer 21

my \[\Delta x\] is 6/n

Answer 22

we can at least break it apart into two pieces, since addition is commutative
and also we need to write
\[\Delta x=\frac{5+1}{n}=\frac{6}{n}\]
\[\lim_{n\rightarrow \infty} \sum_{k=1}^n 1 \Delta x=\lim_{n\rightarrow \infty}\sum_{k=1}^n \frac{6}{n}\]
\[=\frac{6}{n}\sum_{k=1}^n 1=6\]

Answer 23

right we use
\[\Delta x = \frac{6}{n}\]

Answer 24

my fk\[\Delta x\] is 108k/n -12/n

Answer 25

correct or not?

Answer 26

are you breaking it in to two pieces, or just using
\[1+3x\]?

Answer 27

i would make my life easier and write the sum in two parts, since we have already computed the first part. it is 6

Answer 28

then i would look at the function
\[f(x)=x\] on (-1,5) and say
\[x_0=-1, x_1=-1+\frac{6}{n}, x_2=-1+\frac{12}{n} ...x_k=-1+\frac{6k}{n}\]

Answer 29

i realize we have to multiply by 3, but that comes right out front of the summation because
\[\sum 3f(x)=3\sum f(x)\]

Answer 30

so all that is left is to write
\[\sum_{k=1}^n (-1+\frac{6k}{n})\times \frac{6}{n}\]

Answer 31

oh times 3

Answer 32

\[3\sum_{k=1}^n (-1+\frac{6k}{n})\times \frac{6}{n}\]

Answer 33

why times 3?

Answer 34

\[3\sum_{k=1}^n -\frac{6}{n}+\frac{36k}{n^2}\]
\[-18+\frac{36}{n^2}\sum_{k=1}^n k\]

Answer 35

oh because i want to make life easy

Answer 36

we have
\[f(x)=1+3x\] and so i am just saying we can sum up as
\[\sum (1+3x) dx=\sum 1dx +3\sum x dx\]

Answer 37

the first sum we already computed here
\[=\frac{6}{n}\sum_{k=1}^n 1=6\]

Answer 38

and it is just easier for me to work with
\[f(x)=x\] than
\[f(x)=3x\]although it really makes no difference the 3 comes right out front of the summation

Answer 39

however i forgot the distributive law! it should be
\[3\sum_{k=1}^n (-1+\frac{6k}{n})\times \frac{6}{n}\]
\[=3\sum_{k=1}^n -\frac{6}{n}+\frac{36k}{n^2}\]
\[=\frac{-18}{n}\sum_{k=1}^n 1+\frac{108}{n^2}\sum_{k=1}^n k\]
\[=-18+\frac{108}{n^2}\sum_{k=1}^n k\] and only now do we use the summation formula
\[\sum_{k=1}^n k =\frac{n(n+1)}{2}\]

Answer 40

\[\frac{108}{n^2}\frac{n(n+1)}{2}\]
\[\frac{54(n^2+n)}{n^2}\] take the limit as n goes to infinity, get 54

Answer 41

so we should get
\[6-18+54\] as our answer, i hope it is right.. yup,
\[6-18+54=42\]

Answer 42

how to get 54?

Answer 43

start with this line here
\[=-18+\frac{108}{n^2}\sum_{k=1}^n k\] and ignoring the -18 we need to compute
\[\frac{108}{n^2}\sum_{k=1}^n k\]
now we use the summation formula (closed form of the sum) to get
\[\frac{108}{n^2}\times \frac{n(n+1)}{2}\] clear so far?

Answer 44

yes

Answer 45

so we are done with summations completely. we have computed everything. all that is left to do is to take the limit as
\[\Delta x \rightarrow 0\] i.e. as
\[n\rightarrow \infty\]

Answer 46

and if you look at this expression
\[\frac{108}{n^2}\frac{n(n+1)}{2}=\frac{54(n^2+n)}{n^2}\] you want
\[\lim_{n\rightarrow \infty}\frac{54(n^2+n)}{n^2}\]

Answer 47

you have a rational function where the numerator and denominator both have degree 2, so you take the ratio of the leading coefficients and get that limit is 54 pretty much in your head

Answer 48

i suppose you could write something out fancier like divide top and bottom by n^2 but that is a waste of time. you could even use l'hopital if you like but that is like killing a gnat with a 45

Answer 49

if you want to understand further, since we have the answer, instead of breaking it in to two pieces you could keep the sum as
\[\sum f(x_k)\frac{6}{n}=\sum 1+3(-1+\frac{6k}{n})\frac{6}{n}\] but the work will be identical

Answer 50

i get like that

Answer 51

oke2
i try again.

Answer 52

then your next step will be to break it up in to pieces just like we did previous, and to compute just as before, so it will be repetition, but you will learn something by doing it

Answer 53

how get the 6-18+54?

Answer 54

until here,,i just get until 54

Answer 55

where does 6-18+54 come from?

Answer 56

6 from the first sum, -18 from the second part