Question

Find a formula for the partial sum
\[\large 1+4r+9r^2+\cdots+n^2r^{n-1}\]

Answer 1

\[\sum_{i=1}^{n}i^2r^{i-1}\]

Answer 2

What have you tried so far?

Answer 3

I actually found two different solutions, just trying to understand this problem better by looking at how others approach this

Answer 4

\[
f(r) = \sum_{k=0}^{n}r^k=\frac{1-r^{n+1}}{1-r}\\
f'(r) = \sum_{k=1}^{n}kr^{k-1}\\
rf'(r) = \sum_{k=1}^{n}kr^{k}\\
(rf'(r))' = \sum_{k=1}^{n}k^2r^{k-1} = rf''(r) + f'(r)
\]

Answer 5

Wow! that's almost same as the first solution I have :

\[\sum_{i=1}^{n}i^2r^{i-1} =\dfrac{d}{dr} r\sum_{i=1}^{n} \left(r^i\right)' =\dfrac{d}{dr} r\left(\sum_{i=1}^{n} r^i\right)' = \dfrac{d}{dr} r \left(r\dfrac{r^n-1}{r-1}\right)' \]

Answer 6

actually it is identical xD

Answer 7

The second method uses some kind of convolution of two interleaving sequences

Answer 8

\[
f(r) = \sum_{k=0}^{n}r^k
\]And \[
S_n = \sum_{k=1}^nkr^{k-1}
\]Then\[
rS_n = \sum_{k=1}^nkr^{k}
\]Let \(k=m-1\):\[
rS_n = \sum_{m=2}^{n+1}(m-1)r^{m-1} = \sum_{m=2}^{n+1}mr^{m-1} -\sum_{m=2}^{n+1}r^{m-1}
\]Undo the sub for the second term:\[
rS_n= \sum_{m=2}^{n+1}mr^{m-1} -\sum_{k=1}^{n}r^{k}
\]Add missing terms and remove extraneous terms:\[
rS_n+1 - (n+1)r^{n+1} = S_n - f(r)
\]So we get \[
S_n = \frac{1-(n+1)r^{n+1}+f(r)}{1-r}
\]When we do:\[
\Psi _n = \sum_{k=1}^nk^2r^{k-1}
\]We get \[
r\Psi_n = \sum_{m=2}^n(m-1)^2r^{m-1}
\]And we can use previous methods to solve for it.

Answer 9

Looks pretty neat, so if i understand correctly we need to find \(\sum kr^{k-1}\) before finding \(\sum k^2r^{k-1}\)

Answer 10

\[r\Psi_n = \sum_{m=\color{red}{1}}^n(m-1)^2r^{m-1} = \Psi_n -2\sum_{m=\color{red}{1}}^nmr^{m-1} + \sum_{m=\color{red}{1}}^n r^{m-1}\]

Answer 11

we can simply plugin the previous result for the sum in middle, awesome!

Answer 12

\[
f(m,r) = \sum_{k=0}^nk^mr^{k}
\]And \[
rf(m,r) = \sum_{k=0}^nk^mr^{k+1} =\sum_{k'=1}^{n+1}(k'-1)^mr^{k'}
\]\[
rf(m,r)+(-1)^m-n^mr^{n+1}= \sum_{p=0}^{q}{q\choose p}(-1)^pf(q-p,r)
\]This is sort of our recursive solution

Answer 13

Hmmm, I guess to get it explicit:\[
f(m,r) = \frac{(-1)^m-n^mr^{n+1}+\sum_{k=0}^{n-1}{n\choose k}(-1)^pf(q-p,r)}{1-r}
\]

Answer 14

I know I made a small error in it

Answer 15

that was a typo in the exponent, its okay it is a beautiful solution !

Answer 16

\[
f(m,r) = \frac{(-1)^m-n^mr^{n+1}-\sum_{k=0}^{m-1}{m\choose k}(-1)^pf(m-k,r)}{1-r}
\]This one seems to work for \(m=0\).

Answer 17

Hmm, I still shouldn't have that n term, they should be m most likely

Answer 18

what is r here?

Answer 19

I think r could be any real number

Answer 20

\[\large f(m,r) = \frac{1}{1-r}\left[n^m(r^n-1)+\sum\limits_{k=2}^n \sum\limits_{i=0}^{m-1} \binom{m}{i}k^i(r^k-1)\right]\]

Answer 21

The whole point though is to get rid of that outer sum. And you shouldn't have any \(n\) terms.

Answer 22

"n" is the index, nth partial sum right

Answer 23

oh, right, I forgot

Answer 24

I was just messing with your work and ended up wid above formula
it is kind of recursive too as the right side part has one exponent less..

Answer 25

Everything goes to hell when you let \(r=1\) though.

Answer 26

\[
f(m, 1) = \sum_{k=0}^nk^m
\]

Answer 27

Haha lets just say \(r\ne 1\) then

Answer 28

But \(r=1\) is an interesting series.

Answer 29

Indeed, it can be shown that that series asymptotically approaches \(\dfrac{n^{m+1}}{m+1}\)
\[f(m,1) \sim \dfrac{n^{m+1}}{m+1}\]

Answer 30

Is there a formula for that series?

Answer 31

Yes, was working on some definite integral last night and ended up wid above result...
let me pull up..

Answer 32

I think \(f(-1,1)\) is known not to have an elementary function.

Answer 33

Sorry it is not a formula, just a limit which gives the asymptotic sum

Answer 34

\[\lim\limits_{n\to\infty}\dfrac{f(m,1)}{n^{m+1}} = \frac{1}{m+1}\]

Answer 35

for r=1 sum if n(n+1)(2n+1)/6 you can prove using induction

Answer 36

http://math.stackexchange.com/questions/936924/limit-of-the-sequence-frac1k2k-nknk1

Answer 37

right @sparrow2 thats sum of squares of first n natural numbers, fun one!

Answer 38

i think the point is that we just should write the sum in the form so when you give me a natural number for ex n=39 i will give you the answer not the sum written in other form

Answer 39

like in sum of squares of first n

Answer 40

you mean explicit formula

Answer 41

i don't know what it is called

Answer 42

i don't know math's english terminology very well :)

Answer 43

got you :) basically we're trying to find a recursive relation for
\[\large 1^m + 2^mr + 3^mr^2+\cdots+n^mr^{n-1}\]
when \(r=1\), we get the series
\[1^m+2^m+3^m+\cdots+n^m\]

Answer 44

yeah the original question is a special case, \(m=2\)

Answer 45

I am a little late to the party, but I thought I'd try to challenge myself to come up with a different approach. I think it doesn't help much, but is there anything to be gained by writing the squares as:
\[k^2 = \sum_{n=1}^k (2n-1) \]
My thought was to switch the order of summation signs, however it won't work here since the upper limit k is summed over in the other summation, oh well.

Answer 46

Thats similar to what wio did in his recurrence relation,
that works pretty nicely as it reduces the exponent.. .

Answer 47

Oh I thought he just did the derivatives of the geometric series approach.

Answer 48

He did it in two ways..

Answer 49

and do we know that there is a formula for this?

Answer 50

check this @sparrow2
https://en.wikipedia.org/wiki/Faulhaber%27s_formula

Answer 51

ok thanks

Answer 52

generating functions might also work i think
post the work @ikram002p

Answer 53

Let \[f(n,m,r)=\sum\limits_{k=1}^n k^mr^{k-1} \]

Let \(a_k = k^m\) and \(b_k = r^{k-1}\),
then \(B_n=\sum\limits_{k=1}^n b_k = \dfrac{r^n-1}{r-1}\)

It is easy to see that \(b_n = B_{n+1}-B_n\)

\[\begin{align}\color{blue}{f(n,m,r)}&=\sum\limits_{k=1}^n a_k b_k \\~\\
&=\sum\limits_{k=1}^n a_k (B_{n+1}-B_n) \\~\\
&= a_nB_n+\sum\limits_{k=2}^n (a_{k-1}-a_{k}) B_k \\~\\
&= n^m\dfrac{r^n-1}{r-1}+\frac{1}{1-r}\sum\limits_{k=1}^n [(k-1)^m-k^m] (r^k-1) \\~\\

\end{align}\]

Answer 54

Liam did an investigation to see how water temperature affects the amount of salt that will dissolve in the water. He filled 4 beakers with exactly 100 milliliters of water each. He then heated each beaker to a different temperature and tested salt solubility at each different temperature. What was Liam's independent variable?








amount of water in each beaker








the number of beakers








the amount of salt that dissolved in each beaker








the temperature of the water in each beaker

Answer 55

A student conducts an experiment to determine how the additional of salt to water affects the density of the water. The student fills 3 beakers with equal amounts of water. He then adds 1 cup of salt to the first beaker, 2 cups of salt to the second beaker and no salt to the third beaker. What is the dependent variable in the student's investigation?








the amount of salt in the water








the temperature of the water








the density of the water








the ph of the water