Proving $\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$ without induction [duplicate]

Solution 1:

$$n^{3}-(n-1)^{3}=3n^{2}+3n+1$$ $$(n-1)^{3}-(n-2)^{3}=3(n-1)^{2}+3(n-1)+1$$ $$\vdots$$ $$2^{3}-1^{3}=3(1)^{2}+3(1)+1$$

Now use telescopic cancellation.

Here are some "proof without words"(I find them more elegant):

Finally a more generalized form:$$1^{k}+2^{k}+\cdots+n^{k}=\sum\limits_{i=1}^{k}S(k,i)\binom{n+1}{i+1}i!$$ Where S(k,i) represents the Stirling number of the second kind.

Solution 2:

HINT:

To find $\sum_{1\le r\le n}r^m$ we can utilize the identity

$$(r+1)^{m+1}-r^{m+1}$$ $$=\sum_{1\le t\le m+1}\binom {m+1}t r^{m+1-t}=(m+1)r^m+\sum_{2\le t\le m+1}\binom {m+1}t r^{m+1-t}$$ and need to know $\sum_{1\le r\le n}r^s$ for $0\le s\le m-1$

For $m=2,$ $$(r+1)^3-r^3=3r^2+\sum_{2\le t\le 3}\binom 3t r^{3-t}=3r^2+3r+1$$

Put $r=1,2,3,\cdots,n-1,n$ and add to get $$(n+1)^3-1^3=3\sum_{1\le r\le n}r^2+3\sum_{1\le r\le n}r+\sum_{1\le r\le n}1$$

Now, we know $\sum_{1\le r\le n}r=\frac{n(n+1)}2$ and $\sum_{1\le r\le n}1=n$