Proof without using induction How to prove that 
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
without using induction.
If we don't know the right side of this expression, how to get right expression. I tried with partial sums and binomial formula but can't get it.
So the problem is:
$$1^2+2^2+...+n^2=?$$
Thanks for replies.
 A: Evaluate the telescoping sum 
$$\sum_{k=1}^n k^3-(k-1)^3=n^3$$
But $k^3-(k-1)^3=3k^2-3k-1$.  Thus, 
$$\sum_{k=1}^n k^3-(k-1)^3=n^3=\sum_{k=1}^n (3k^2-3k-1)$$
Isolate the sum over $k^2$ and use the result of the sum of an arithmetic sum and you will have it!
This methodology, using a telescoping sum, works for higher powers.  So, it can be used to find the sum of $k^m$ for any integer $m$.
A: Assume we know $\sum\limits_{k=1}^nk=\frac{n(n+1)}{2}$. Compute the following cubes
$$\begin{align}
1^3&=1\\
(1+1)^3&=1^3+3\cdot 1^2+3\cdot 1+1^3\\
\vdots&=\vdots\\
n^3&=(n-1)^3+3(n-1)^2+3(n-1)+1^3\\
(n+1)^3&=n^3+3n^2+3n+1^3\end {align}$$
Add these equations together and cancel the cubes you have on both sides you get
$$\begin{align}(n+1)^3&=1+3\sum_{k=1}^nk^2+3\sum_{k=1}^nk+n\\
&=(n+1)\frac{3n+2}{2}+3\sum_{k=1}^nk^2\end{align}$$
This yields 
$$\sum_{k=1}^nk^2=\frac{n+1}{3}\left(n^2+2n+1-\frac{3n+2}{2}\right)=\frac{(n+1)(2n^2+n)}{6}$$
Factoring $n$ we get the result expected
A: $\sum_ 1^{n+1}k^2=\sum_ 0^{n}(k+1)^2=\sum _0^{n}[k^2+2k+1]=\sum_1^nk^2+2\cdot \sum _1^nk + (n+1)$ which gives $(n+1)^2=2\cdot \sum_1^n k + (n+1)$ from which you deduce that $\sum_1^nk=\frac {(n+1)^2-(n+1)}{2}=\frac{(n+1)n}{2}$, the familiar formula. Now you can do the same but start with $\sum_1^{n+1}k^3=\sum_0^n(k+1)^3=\cdots $ and with a bit more work but essentially the same kind of algebra, and using the formula for $\sum_1^nk $ which was just derived you discover the formula without guessing. This will work for any specified power. It should be noted though that this still uses induction, hidden in the manipulation of sums. 
A: Seven or eight ways to do this are presented in Concrete Mathematics by Knuth et. al.  The most elementary is this:
Let $S_r(n) = \sum_{k=0}^n x^r$; we are trying to find a closed form for $S_2(n)$.
Pretend we were trying to find $S_3(n)$:
$$
S_3(n)  + (n+1)^3 = S_3(n+1) = \sum_{k=0}^n (k+1)^3 = \sum_{k=0}^n (j^3+3k^2+3k+1) \\
= \sum_{k=0}^n k^3 + 3\sum_{k=0}^n k^2  + 3 \sum_{k=0}^n k + \sum_{k=0}^n 1
= S_3(n) + 3S_2(n)  + 3 \frac{n(n+1)}{2} + n
$$
The $S_3(n)$ terms drop out (which would be harsh if we were really trying to find  $S_3(n)$ ) and we get
$$
3S_2(n) = (n+1)^3 -  3\frac{n(n+1)}{2} - n
$$
from which easy algebra gives you your closed form for $S_2(n)$.
A: With linear algebra:
You can do it with linear algebra, by noticing that the map $\Sigma$ which maps a sequence $u_n$ to the sequence $u_0 + \dots + u_n$ has, as a left inverse, the map $\Delta$, mapping the sequence $u_n$ to the sequence $u_n - u_{n-1}$; since, for any sequence $u_n = f(n)$ where $f$ is a polynomial of degree $d$, $\Delta u$ is polynomial of degree $d-1$, we see that $\Sigma$ maps polynomials of degree $d$ to polynomials of degree $d+1$, and we can find which ones by linear algebra.
With combinatorics:
An even simpler way is to use the standard basis $e_d$ where $e_d$ is the sequence $e_{d,n} = \binom{n+d}{d}$, for then we have
$$(\Delta e_d)_n = e_{d,n} - e_{d,n-1} = \binom{n+d}{d} - \binom{n+d-1}{d} = \binom{n+d-1}{d-1} = (e_{d-1})_{n}.$$
Then we only need to write $n^2$ as a linear combination of $e_{0,n} = 1$, $e_{1,n} = n$ and $e_{2,n} = n(n-1)/2$.
With integrals:
A third way is to use integrals: namely, with simple linear algebra we find that
$$ n^2 = \int_{n-1}^{n} (x^2+x+1/6) dx,$$
so that
$$ 0+1+\dots+n^2 = \int_0^{n} (x^2+x+1/6) dx = \frac{1}{6} (2n^3+3n^2+n).$$
With geometry:
A fourth way is the usual geometric proof, where you can tile together six pyramids in a parallelepiped.
A: Any statement on natural numbers can be proved only using induction. It is possible that the induction is hidden in the argument. So this question doesn't make sense.
