$n^2 + (n-1)^2 + \dots +2^2 + 1^2$ equals what?

Question

$$n^2 + (n-1)^2 + \dots +2^2 + 1^2 = \text{??}$$

I am reading that this equals

$$\frac{1}{3} n \left( n + \frac{1}{2} \right) (n+1) $$

But have no clue how.. The thing that strikes me most is the fact that the latter has a cubed factor when expanded (i.e. in big O notation it's O$(n^3)$).

What happens to the final formula if you add term $(n+1)^2$ to it? — abiessu, Sep 13 '13 at 12:25
Have you tried Induction??I meant try for first few $n$ and then generalise. — Vishesh, Sep 13 '13 at 12:25
It should not be surprising that it becomes a polynomial of degree $3$. After all, you have $n$ terms that are each of size approximately $n^2$ (ok, very approximately for the small ones), which should give you $n^3$ in total. — Tobias Kildetoft, Sep 13 '13 at 12:31
The fact that the sum $S(n)$ is cubic in $n$ is the least surprising thing, since the difference $S(n)-S(n-1)=n^2$ is quadratic. — Andrea Mori, Sep 13 '13 at 12:32
since the sum $1 + 2 + \dots + n$ is O$(n^2)$, it should be no big surprise that this formula could be O$(n^3)$. — mau, Sep 13 '13 at 12:33
It is more surprising that the expression for $S(n)$ contains denominators, yet gives an integer result for every integer value of $n$. — Andrea Mori, Sep 13 '13 at 12:33
You should be able to find a few questions about the same sum by looking at frequent summation questions. — Martin Sleziak, Sep 13 '13 at 15:57

score 8 · Answer 1 · answered Sep 13 '13 at 12:46

8

This can be done in several ways (and has been done many times on this site). Let's try something a bit different.

Let $$f(x)=\frac13x(x+\frac12)(x+1).$$ Then by differentiating we get $$ g(x)=f'(x)=x^2+x+\frac16. $$ It is easy to verify that $$ g(x)-g(x-1)=2x. $$ Integrating this gives us $$ f(x)-f(x-1)=x^2+C $$ for some constant $C$. As $f(0)=f(-1)=0$, we see that $C=0$, so for all $n$ we have $$ f(n)-f(n-1)=n^2 $$ Thus we are at the point Andrea Mori described. The rest is easy $$ 1^2+2^2+\cdots+n^2=(f(1)-f(0))+(f(2)-f(1))+\cdots+(f(n)-f(n-1))=f(n)-f(0)=f(n). $$

answered Sep 13 '13 at 12:46

Jyrki Lahtonen

140,891

It may be even easier to repeat the dose, and differentiate $g(x)$ also. So $g'(x)=2x+1$ and thus $g'(x)-g'(x-1)=2$. Therefore $g(x)-g(x-1)=2x+C$, but $g(0)=g(-1)=1/6$, so... You can also find formulas for $\sum_k k^3$ and higher powers in this way by integrating $f(x)$. Finding the constants of integration brings about Bernoulli numbers. – Jyrki Lahtonen Sep 13 '13 at 12:51
I made this CW, because I have a recollection of having described the techique in an earlier answer. Can't locate it now. – Jyrki Lahtonen Sep 13 '13 at 12:56

score 3 · Accepted Answer · answered Sep 13 '13 at 12:46

For a geometric "proof" or explanation with respect to how to make sense of the resulting identity for the sum of n consecutive squares, see the entry here.

It uses the equivalent identity: $$1^2 + 2^2 + \cdots + (n - 1)^2 + n^2 = \sum_{k = 1}^n k^2 = \frac 13(n)\left(n + \frac 12\right)(n+1) =\dfrac {n(n+1)(2n+1)}{6} $$ which can easily be "proven" by induction on $n$.

Andrea Mori · Answer 3 · 2013-09-13T13:12:12.253

Let $$ S(n)=1^2+2^2+\cdots+n^2. $$ Since $S(n)-S(n-1)=n^2$ one can suspect that $S(n)$ is a cubic polynomial in $n$. Obviously, $S(n)\in\Bbb Z$ for all $n=1$, $2$, $3\dots$ and one knows that the ring of numeric polynomals, i.e. the polynomials in $\Bbb Q[X]$ which take integer values on $\Bbb N$ is freely generated over $\Bbb Z$ by the binomial coefficients $$ \binom Xk=\frac1{k!}X(X-1)\cdots(X-k+1). $$ Thus one may attempt to write $$ S(n)=a\binom X0+b\binom n1+c\binom n2+d\binom n3, $$ for some $a,b,c,d\in\Bbb Z$ (since $\binom Xk$ has degree $k$, those with $k\geq4$ will not enter). Some fiddling allows to find the right coefficients and confirm the formula.

score 1 · Answer 4 · answered Sep 13 '13 at 12:48

1

Hint: Use induction on $n$

$$n=1:\quad1^2=\frac{1}{3}(1)\left(\frac{3}{2}\right)(2)=1$$

Assume statement is true for $n$, then \begin{align*} (n+1)^2+n^2+\dots+2^2+1^2&=n^2+2n+1+\frac{1}{3}(n)\left(n+\frac{1}{2}\right)(n+1)\\ \end{align*} Then expand and factorise and you'll get your answer.

answered Sep 13 '13 at 12:48

BlackAdder

4,059

1

Given the expression, it is easy to prove (by induction) that it is true. But how could one come with that expression given the sum of squares? – onimoni Sep 13 '13 at 12:50
3

The question concerns how the closed form for the sum arises in the first place, and how to make sense of it, not about confirming/proving that it's true by induction. – amWhy Sep 13 '13 at 12:50

$n^2 + (n-1)^2 + \dots +2^2 + 1^2$ equals what?

4 Answers4

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