Is there a simplification of this sum of quadratic sums?

Question

I have $a(1+2+3+...+N)^2+a(2+3+...+N)^2+a(3+..+N)^2+...+aN^2$, essentially a sum of quadratic terms holding an integer summation that get consecutively smaller Within each term the summation up to N elements (which could also go to infinity if that makes life easier). Some experimentation lead me to believe that this converge to something when divided by $N^{-5}$, but don't take my word for it. Any simplification here possible?

Answer 1

Mathematica gives the following result: $$ \frac{1}{30} a \; n (n+1) (2 n+1) (2n^2+2n+1) $$ Hopefully I'm interpreting your summation correctly.

It seems to give the correct result if for example letting $N=3$ for your summation gives: $$ a((1+2+3)^2+(2+3)^2+(3)^2))=70a $$ which is what the formula above gives.

EDIT:

This is a rather messy prove but it does seem to work and is relatively straightforward.

Looking at it a bit more you could get the result above by writing you summation as: $$ a \sum _{j=1}^n \left(\sum _{i=j}^n i\right){}^2 $$ then writing the inner sum from $i=j$ to $i=n$ as: $$ \frac{1}{2} (n-j+1) (j+n) \;\; \text{ which after substituting gives } \;\;a \sum _{j=1}^n[\frac{1}{2} (n-j+1) (j+n)]^2 $$ $$= \frac{a}{4} \sum _{j=1}^n[(n-j+1) (j+n)]^2 $$ If you then expand $[(n-j+1) (j+n)]^2 $ and then collect the terms of $j$ together since that is the index of the outer most summation. Which gives a rather messy result: $$j^4-2 j^3+j^2 (-2 n^2-2 n+1)+j (2 n^2+2 n)+n^4+2 n^3+n^2$$ But then all you need to know are the summations the powers up to $4$. Since we can then replace the above expression for $ \left(\sum _{i=j}^n i\right){}^2 $ in the outer summation again (which is indexed from $j=1$ to $n$) gives. $$\sum _{j=1}^n[j^4]-2 \sum _{j=1}^n[j^3]+ \left(-2 n^2-2 n+1\right)\sum _{j=1}^n[j^2]+ \left(2 n^2+2 n\right)\sum _{j=1}^n[j]+ (n^4+2 n^3+n^2)\sum _{j=1}^n[1]$$

Which after substituting in the known formula's for the sums of $j$, $j^2$, $j^3$ and $ j^4$ from to 1 to $n$ and multiplying by $a/4$ gives the final result $$\frac{1}{30} a\; n (n+1) (2 n+1) (2n^2+2n+1)$$

It's definitely not a pretty proof but it works and is quite straightforward, just a bit messy!

Answer 2

Drop the $a$ for now.

Then, the $n$th term is, \begin{align}a_n&=\left(\displaystyle\sum_{i=n}^{N} i\right)^2 \\&=\left(\dfrac12N(N+1)-\dfrac12n(n-1)\right)^2 \\&=\dfrac14(N^2(N+1)^2+n^2(n-1)^2-2Nn(N+1)(n-1)).\end{align}

So, \begin{align} \displaystyle\sum_{n=1}^N a_n &=\frac14\sum_{n=1}^N N^2(N+1)^2+\frac14\sum_{n=1}^N n^2(n-1)^2-\frac14\sum_{n=1}^N 2Nn(N+1)(n-1) \\&=\frac14N^3(N+1)^2+\frac14\sum_{n=1}^N n^2(n-1)^2-\frac12N(N+1)\sum_{n=1}^N n(n-1).\end{align}

Provided you know the closed-forms of $\sum_{n=1}^N n^k$ for $k=1,2,3,4$, you should be good!

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If you can guess that the answer will be a degree $5$ polynomial (and it's not hard since each term is a degree $4$ polynomial), then another approach would be to first guess the closed-form to be $aN^5+bN^4+cN^3+dN^2+eN$.

Then for $N=1$, it should equal $1$.

So, we get $a+b+c+d+e=1$.

Similarly, for $N=2$, it should equal $13$ and thus, $32a+16b+8c+4d+2e=13$.

After getting $5$ (nasty) linear equations, you can solve them to obtain your answer.

Admittedly, this is yet another cumbersome approach but I thought I'd mention this since as you've already mentioned in your question, you already know that the closed-form will have degree $5$.

Answer 3

Factor $a$ out. Let $1+2+ \cdots + n = \frac{n(n+1)}2 = b$. Then, we have

$b^2+(b-1)^2+ \cdots + (b- n(n-1)/2)^2 = $

$(n-1)b^2-2b(1+ \cdots + n(n-1)) +1^2+2^2+ \cdots + (n(n-1)/2)^2$.

All of these sums have well known closed forms so I will leave the rest to you.