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ganeshie8
 one year ago
Find a formula for the partial sum
\[\large 1+4r+9r^2+\cdots+n^2r^{n1}\]
ganeshie8
 one year ago
Find a formula for the partial sum \[\large 1+4r+9r^2+\cdots+n^2r^{n1}\]

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UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{i=1}^{n}i^2r^{i1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What have you tried so far?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I actually found two different solutions, just trying to understand this problem better by looking at how others approach this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(r) = \sum_{k=0}^{n}r^k=\frac{1r^{n+1}}{1r}\\ f'(r) = \sum_{k=1}^{n}kr^{k1}\\ rf'(r) = \sum_{k=1}^{n}kr^{k}\\ (rf'(r))' = \sum_{k=1}^{n}k^2r^{k1} = rf''(r) + f'(r) \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Wow! that's almost same as the first solution I have : \[\sum_{i=1}^{n}i^2r^{i1} =\dfrac{d}{dr} r\sum_{i=1}^{n} \left(r^i\right)' =\dfrac{d}{dr} r\left(\sum_{i=1}^{n} r^i\right)' = \dfrac{d}{dr} r \left(r\dfrac{r^n1}{r1}\right)' \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2actually it is identical xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2The second method uses some kind of convolution of two interleaving sequences

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(r) = \sum_{k=0}^{n}r^k \]And \[ S_n = \sum_{k=1}^nkr^{k1} \]Then\[ rS_n = \sum_{k=1}^nkr^{k} \]Let \(k=m1\):\[ rS_n = \sum_{m=2}^{n+1}(m1)r^{m1} = \sum_{m=2}^{n+1}mr^{m1} \sum_{m=2}^{n+1}r^{m1} \]Undo the sub for the second term:\[ rS_n= \sum_{m=2}^{n+1}mr^{m1} \sum_{k=1}^{n}r^{k} \]Add missing terms and remove extraneous terms:\[ rS_n+1  (n+1)r^{n+1} = S_n  f(r) \]So we get \[ S_n = \frac{1(n+1)r^{n+1}+f(r)}{1r} \]When we do:\[ \Psi _n = \sum_{k=1}^nk^2r^{k1} \]We get \[ r\Psi_n = \sum_{m=2}^n(m1)^2r^{m1} \]And we can use previous methods to solve for it.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Looks pretty neat, so if i understand correctly we need to find \(\sum kr^{k1}\) before finding \(\sum k^2r^{k1}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[r\Psi_n = \sum_{m=\color{red}{1}}^n(m1)^2r^{m1} = \Psi_n 2\sum_{m=\color{red}{1}}^nmr^{m1} + \sum_{m=\color{red}{1}}^n r^{m1}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we can simply plugin the previous result for the sum in middle, awesome!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(m,r) = \sum_{k=0}^nk^mr^{k} \]And \[ rf(m,r) = \sum_{k=0}^nk^mr^{k+1} =\sum_{k'=1}^{n+1}(k'1)^mr^{k'} \]\[ rf(m,r)+(1)^mn^mr^{n+1}= \sum_{p=0}^{q}{q\choose p}(1)^pf(qp,r) \]This is sort of our recursive solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm, I guess to get it explicit:\[ f(m,r) = \frac{(1)^mn^mr^{n+1}+\sum_{k=0}^{n1}{n\choose k}(1)^pf(qp,r)}{1r} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know I made a small error in it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that was a typo in the exponent, its okay it is a beautiful solution !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(m,r) = \frac{(1)^mn^mr^{n+1}\sum_{k=0}^{m1}{m\choose k}(1)^pf(mk,r)}{1r} \]This one seems to work for \(m=0\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, I still shouldn't have that n term, they should be m most likely

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I think r could be any real number

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\large f(m,r) = \frac{1}{1r}\left[n^m(r^n1)+\sum\limits_{k=2}^n \sum\limits_{i=0}^{m1} \binom{m}{i}k^i(r^k1)\right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The whole point though is to get rid of that outer sum. And you shouldn't have any \(n\) terms.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2"n" is the index, nth partial sum right

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I 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..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Everything goes to hell when you let \(r=1\) though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(m, 1) = \sum_{k=0}^nk^m \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Haha lets just say \(r\ne 1\) then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But \(r=1\) is an interesting series.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Indeed, it can be shown that that series asymptotically approaches \(\dfrac{n^{m+1}}{m+1}\) \[f(m,1) \sim \dfrac{n^{m+1}}{m+1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there a formula for that series?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, was working on some definite integral last night and ended up wid above result... let me pull up..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think \(f(1,1)\) is known not to have an elementary function.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Sorry it is not a formula, just a limit which gives the asymptotic sum

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim\limits_{n\to\infty}\dfrac{f(m,1)}{n^{m+1}} = \frac{1}{m+1}\]

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.0for r=1 sum if n(n+1)(2n+1)/6 you can prove using induction

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2http://math.stackexchange.com/questions/936924/limitofthesequencefrac1k2knknk1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2right @sparrow2 thats sum of squares of first n natural numbers, fun one!

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.0i 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

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.0like in sum of squares of first n

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2you mean explicit formula

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.0i don't know what it is called

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.0i don't know math's english terminology very well :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2got you :) basically we're trying to find a recursive relation for \[\large 1^m + 2^mr + 3^mr^2+\cdots+n^mr^{n1}\] when \(r=1\), we get the series \[1^m+2^m+3^m+\cdots+n^m\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2yeah the original question is a special case, \(m=2\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I 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: \[k^2 = \sum_{n=1}^k (2n1) \] 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Thats similar to what wio did in his recurrence relation, that works pretty nicely as it reduces the exponent.. .

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Oh I thought he just did the derivatives of the geometric series approach.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2He did it in two ways..

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.0and do we know that there is a formula for this?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2check this @sparrow2 https://en.wikipedia.org/wiki/Faulhaber%27s_formula

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2generating functions might also work i think post the work @ikram002p

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Let \[f(n,m,r)=\sum\limits_{k=1}^n k^mr^{k1} \] Let \(a_k = k^m\) and \(b_k = r^{k1}\), then \(B_n=\sum\limits_{k=1}^n b_k = \dfrac{r^n1}{r1}\) It is easy to see that \(b_n = B_{n+1}B_n\) \[\begin{align}\color{blue}{f(n,m,r)}&=\sum\limits_{k=1}^n a_k b_k \\~\\ &=\sum\limits_{k=1}^n a_k (B_{n+1}B_n) \\~\\ &= a_nB_n+\sum\limits_{k=2}^n (a_{k1}a_{k}) B_k \\~\\ &= n^m\dfrac{r^n1}{r1}+\frac{1}{1r}\sum\limits_{k=1}^n [(k1)^mk^m] (r^k1) \\~\\ \end{align}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Liam 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A 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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