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

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

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\[\sum_{i=1}^{n}i^2r^{i-1}\]
What have you tried so far?
I actually found two different solutions, just trying to understand this problem better by looking at how others approach this

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\[ f(r) = \sum_{k=0}^{n}r^k=\frac{1-r^{n+1}}{1-r}\\ f'(r) = \sum_{k=1}^{n}kr^{k-1}\\ rf'(r) = \sum_{k=1}^{n}kr^{k}\\ (rf'(r))' = \sum_{k=1}^{n}k^2r^{k-1} = rf''(r) + f'(r) \]
Wow! that's almost same as the first solution I have : \[\sum_{i=1}^{n}i^2r^{i-1} =\dfrac{d}{dr} r\sum_{i=1}^{n} \left(r^i\right)' =\dfrac{d}{dr} r\left(\sum_{i=1}^{n} r^i\right)' = \dfrac{d}{dr} r \left(r\dfrac{r^n-1}{r-1}\right)' \]
actually it is identical xD
The second method uses some kind of convolution of two interleaving sequences
\[ f(r) = \sum_{k=0}^{n}r^k \]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.
Looks pretty neat, so if i understand correctly we need to find \(\sum kr^{k-1}\) before finding \(\sum k^2r^{k-1}\)
\[r\Psi_n = \sum_{m=\color{red}{1}}^n(m-1)^2r^{m-1} = \Psi_n -2\sum_{m=\color{red}{1}}^nmr^{m-1} + \sum_{m=\color{red}{1}}^n r^{m-1}\]
we can simply plugin the previous result for the sum in middle, awesome!
\[ f(m,r) = \sum_{k=0}^nk^mr^{k} \]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
Hmmm, I guess to get it explicit:\[ f(m,r) = \frac{(-1)^m-n^mr^{n+1}+\sum_{k=0}^{n-1}{n\choose k}(-1)^pf(q-p,r)}{1-r} \]
I know I made a small error in it
that was a typo in the exponent, its okay it is a beautiful solution !
\[ f(m,r) = \frac{(-1)^m-n^mr^{n+1}-\sum_{k=0}^{m-1}{m\choose k}(-1)^pf(m-k,r)}{1-r} \]This one seems to work for \(m=0\).
Hmm, I still shouldn't have that n term, they should be m most likely
what is r here?
I think r could be any real number
\[\large f(m,r) = \frac{1}{1-r}\left[n^m(r^n-1)+\sum\limits_{k=2}^n \sum\limits_{i=0}^{m-1} \binom{m}{i}k^i(r^k-1)\right]\]
The whole point though is to get rid of that outer sum. And you shouldn't have any \(n\) terms.
"n" is the index, nth partial sum right
oh, right, I forgot
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..
Everything goes to hell when you let \(r=1\) though.
\[ f(m, 1) = \sum_{k=0}^nk^m \]
Haha lets just say \(r\ne 1\) then
But \(r=1\) is an interesting series.
Indeed, it can be shown that that series asymptotically approaches \(\dfrac{n^{m+1}}{m+1}\) \[f(m,1) \sim \dfrac{n^{m+1}}{m+1}\]
Is there a formula for that series?
Yes, was working on some definite integral last night and ended up wid above result... let me pull up..
I think \(f(-1,1)\) is known not to have an elementary function.
Sorry it is not a formula, just a limit which gives the asymptotic sum
\[\lim\limits_{n\to\infty}\dfrac{f(m,1)}{n^{m+1}} = \frac{1}{m+1}\]
for r=1 sum if n(n+1)(2n+1)/6 you can prove using induction
http://math.stackexchange.com/questions/936924/limit-of-the-sequence-frac1k2k-nknk1
right @sparrow2 thats sum of squares of first n natural numbers, fun one!
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
like in sum of squares of first n
you mean explicit formula
i don't know what it is called
i don't know math's english terminology very well :)
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 \[1^m+2^m+3^m+\cdots+n^m\]
yeah the original question is a special case, \(m=2\)
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.
Thats similar to what wio did in his recurrence relation, that works pretty nicely as it reduces the exponent.. .
Oh I thought he just did the derivatives of the geometric series approach.
He did it in two ways..
and do we know that there is a formula for this?
check this @sparrow2 https://en.wikipedia.org/wiki/Faulhaber%27s_formula
ok thanks
generating functions might also work i think post the work @ikram002p
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 \(b_n = B_{n+1}-B_n\) \[\begin{align}\color{blue}{f(n,m,r)}&=\sum\limits_{k=1}^n a_k b_k \\~\\ &=\sum\limits_{k=1}^n a_k (B_{n+1}-B_n) \\~\\ &= a_nB_n+\sum\limits_{k=2}^n (a_{k-1}-a_{k}) B_k \\~\\ &= n^m\dfrac{r^n-1}{r-1}+\frac{1}{1-r}\sum\limits_{k=1}^n [(k-1)^m-k^m] (r^k-1) \\~\\ \end{align}\]
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
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

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