$S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.

when, $n=2$




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    $\begingroup$ There is a fairly simple general formula for $S_n$ in general: $$\frac{n(n+1)(2n+1)}{6}$$ $\endgroup$ – Alex Becker Aug 30 '12 at 6:22
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    $\begingroup$ To adhere to what is in your title strictly (so ignoring the information that $S_n$ is the sum of squares of something), how do you expect to find $S_1$ from your relation? More generally trying to find $S_n$ will block when you need $S_d$ where $d$ is the largest odd divisor of $n$ (possibly $n$ itself), for which you get no clue. $\endgroup$ – Marc van Leeuwen Aug 30 '12 at 9:53
  • $\begingroup$ when i was trying to find the sum of the squares of first $n$ natural numbers then i reached a stage where i got the expression $S_{2n}+4S_{n}=n(2n+1)^2$ $\endgroup$ – HOLYBIBLETHE Aug 30 '12 at 10:48
  • $\begingroup$ @Marc van Leeuwen: the title does not permit long sentences, therefore i was unable to provide the entire information in the title. $\endgroup$ – HOLYBIBLETHE Aug 30 '12 at 10:51
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    $\begingroup$ @RajeshKSingh: Actually I was referring to something that is in the title (though not confirmed in the body of the question), namely "with just this one expression", which appears to forbid using other information. Why did you put that in, if you do want to allow using more information? $\endgroup$ – Marc van Leeuwen Aug 30 '12 at 11:41

Here is a closed form solution to your recurrence relation obtained by Maple,

$$ s(n)={n}^{2}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}}s \left( 1 \right) +\frac{{n}^{3}}{3}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}} \left( \left( -1 \right)^{{\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right)^{-1}+\frac{{n}^{2}}{2}{n}^{{\frac {i\pi }{ \ln \left( 2 \right) }}} \left( \left( -1 \right) ^{{\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right) ^{-1}+\frac{1}{6}\,n{n}^ {{\frac {i\pi }{\ln \left( 2 \right) }}} \left( \left( -1 \right) ^{ {\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right) ^{-1 }-{n}^{2}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}} \,$$

Here is a more compact form

$$ s(n) = \left( {n}^{2}\cos \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) +i{n}^{2}\sin \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) \right) s \left( 1 \right) -{n}^{2}\cos \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) +\frac{{n}^{3}}{3}+\frac{{n}^{2}}{2} +\frac{n}{6}-i{n}^{2}\sin \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) \,.$$

where $s(1)$ is your initial condition. If you plug in $s(1)=1$ in the above formula you get the simple formula, just as it has been mentioned in the comments,

$$ \frac{n}{6} \left( n+1 \right) \left( 2\,n+1 \right) \,,$$

which is equal to $ \sum_{i=1}^{n} i^2 $.


If you are interested only in finding sums of the form $ \sum_{i=1}^{n} i^m \,, m=1,2,3,\dots $, then they are simple techniques to find them. See here.

  • 2
    $\begingroup$ I don't understand how a recurrence relation like this could possibly have a solution that only depends on $S_1$. With this recurrence relation, we could freely choose $S_1, S_3, S_5, \ldots$ to take any values and then the relation would provide us values for $S_{2n}$. $\endgroup$ – alex.jordan Aug 30 '12 at 17:14
  • $\begingroup$ @alex.jordan: I had the same kind of puzzlement, but maybe Maple was looking for a solution given by a single expression, and valid for all real (or complex) values of $n$? $\endgroup$ – Marc van Leeuwen Sep 1 '12 at 13:30

Let $S_n=an^3+bn^2+cn+d$ where $a,b,c,d$ are rational numbers.

So, $S_{2n}+4S_n=n^3 12a+n^2 8b + n 6c+5d$

$\implies n^3 12a+n^2 8b + n 6c+5d= n(2n+1)^2=4n^3+4n^2+n$

Comparing the coefficients of the different powers on $n$,


So, $6S_n=2n^3+3n^2+n=n(n+1)(2n+1)\implies S_n=\frac{n(n+1)(2n+1)}{6}$

Also, $S_n -S_{n-1}=n^2\implies S_n=n^2+S_{n-1}=\sum_{1≤r≤n}r^2+S_0=\sum_{1≤r≤n}r^2$

Observe that, we don't need to know the nature or formula of $S_n$. Solution of any such difference equation of any positive integer degree can be attempted this way.

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    $\begingroup$ How do you know that $S_n$ would be a polynomial in $n$, is it that obvious ? and also why of degree $3$ ? $\endgroup$ – pritam Aug 30 '12 at 6:49
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    $\begingroup$ @pritam Because the RHS of your equation is polynomial of degree three; if $S_n$ is polynomial in $n$ of degree three, then $S_{2n}$ is also a degree-three polynomial in $n$, and so too is their sum. You need to work through the algebra to confirm that such a solution does in fact exist, but it's an excellent starting hypothesis. $\endgroup$ – Steven Stadnicki Aug 30 '12 at 7:11
  • $\begingroup$ @pritam: First difference $S_{n+1}-S_{n}$ is a polynomial of degree $2$. So by a general theorem in the calculus of finite differences, $S_n$ is a polynomial of degree $3$. $\endgroup$ – André Nicolas Aug 30 '12 at 7:13
  • $\begingroup$ But isn't it $S_{2n}$ and $S_n$ that are in question? $\endgroup$ – Tunococ Aug 31 '12 at 7:03
  • $\begingroup$ @Tunococ It is. See my comment on alex.jordan's answer. $\endgroup$ – Did Aug 31 '12 at 9:21

With the additional information you provided about $S_n$ (the sum of squares of the first $n$ integers), there's a neat solution (among other solutions) that uses the perturbation method described in Concrete Mathematics.

Let $C_n$ denote the sum of cubes of the first $n$ natural numbers. Then

\begin{equation} \begin{split} C_{n+1} =& C_n + (n+1)^3 = \sum_{k=1}^{n+1}k^3 = \sum_{k=0}^{n}(k+1)^3 = \sum_{k=0}^{n} k^3 + 3k^2 + 3k + 1 \\ =&\sum_{k=0}^{n}k^3 + 3\sum_{k=0}^{n}k^2 + 3\sum_{k=0}^{n}k + \sum_{k=0}^{n}1 = C_n + 3S_n + 3\dfrac{n(n+1)}{2} + (n+1). \end{split} \end{equation}

Hence \begin{equation} \begin{split} S_n =& \dfrac{(n+1)^3}{3} - \dfrac{n(n+1)}{2} - \dfrac{n+1}{3} = \dfrac{(n+1)(2(n+1)^2 - 3n - 2)}{6} \\ =& \dfrac{(n+1)(2n^2+n)}{6} = \dfrac{n(n+1)(2n+1)}{6}. \end{split} \end{equation}



$S_{2n}=S_{n}+S_{(n+1,2n)}$ -------(A)







$S_{(n+1,2n)}=S_{n}+2n^3+n^2$ -------(B)

we, have from (A) and (B),



we, now have,








You could use induction.

Assuming that $$S_{2n}+4S_n=n(2n+1)^2$$ then add terms to both sides so that the left side increments its index: $$ \begin{align} &S_{2n}+4S_n+(2n+1)^2+(2n+2)^2+4(n+1)^2\\ &=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\ S_{2(n+1)}+4S_{n+1}&=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\ &=(n+1)(2n+1)^2+4(n+1)^2+4(n+1)^2\\ &=(n+1)(2n+1)^2+8(n+1)^2\\ &=(n+1)[(2n+1)^2+8(n+1)]\\ &=(n+1)[4n^2+4n+1+8n+8]\\ &=(n+1)[4n^2+12n+9]\\ &=(n+1)(2n+3)^2\\ &=(n+1)(2(n+1)+1)^2\\ \end{align}$$

The base case is established in your question.

  • $\begingroup$ It is not stated clearly in the question body, but from the title the equation $S_{2n}+4S_n=n(2n+1)^2$ is known to OP. The question was not to prove it. $\endgroup$ – Marc van Leeuwen Aug 30 '12 at 9:47
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    $\begingroup$ @Marc Oh right! Well, OP has no chance of deriving the sum of squares formula just from this relation. If this relation is all that there is to go on, then the values of $S_{2n+1}$ can be set to anything and all that the relation will provide is values of $S_{2n}$. $\endgroup$ – alex.jordan Aug 30 '12 at 17:06
  • $\begingroup$ You are absolutely right. In fact the question the OP wants to be solved seems to be as follows. Assume that (1.) $S_n=1^2+2^2+\cdots+n^2$ for every $n$ and that (2.) $S_{2n}+4S_n=n(2n+1)^2$ for every $n$, show that (3.) $S_n=\frac16n(n+1)(2n+1)$ for every $n$. (Of course, (1.) implies (2.) hence all this is rather odd but...) $\endgroup$ – Did Aug 31 '12 at 6:23

Hint $\rm\quad S_n =\, \sum c_k n^k\ \Rightarrow\ S_{2n} + 4\, S_n\, =\: \sum\ (2^k\!+\!4)\ c_k\ =\ 4\, n^3+4\,n^2 + n,\:$ therefore

$$\rm S_n\, =\ \frac{4}{2^{\color{#C00}3}\!+\!4} n^\color{#C00}3 +\frac{4}{2^\color{#0A0}2\!+\!4}n^{\color{#0A0}2} + \frac{1}{2^\color{brown}1\!+\!4} n^{\color{brown}1}\ =\ \frac{n^3}3+\frac{n^3}2 + \frac{n}6\ =\ \frac{n(n+1)(2n+1)}6$$


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