$1^3+2^3+...+n^3= $? How I can find formula

Question

$1^3+2^3+...+n^3=$ has a formula which $(1+2+3+...+n)^2 $ or $[n(n + 1)/2]^2$ . We can verify with ınduction and I know how I can prove it. How we find what is the formula...

I tried like this $1^3+2^3+...+n^3= An^4+ Bn^3+Cn^2+Dn+E$ after a long process i found it this way $A=-299/54,$ $B=-4/9,$ $C=107/54,$ $D=4.$ And then I tried how can show that But I didn't find Maybe The value of a,b,c,d is wrong Can I get proof of this formula in this way or another way if we didn't know the formula, how would we get this formula

Answer 1

Coming up with a formula is often a matter of 'playing around' ... and hopefully you start to see any patterns.

So, we could just see what happens:

\begin{array}{c|c} n&1^3+2^3+ ... +n^3\\ \hline 1&1\\ 2&9\\ 3&36\\ 4&100\\ ...&...\\ \end{array}

Well ... if you know a little about numbers, I think you'd recognize $1,9,36,100$ as $1^2,3^2,6^2$ and $10^2$ respectively ... and the $1,3,6,10,...$ clearly follows a pattern as well: $1$, $1+2$, $1+2+3$, ...

OK, so there's your formula.

Of course, at this point it's just a hypothesis ... but when you try to prove it ... you find that you can!

Another method is to play around with the veru concepts involved. So, you could take a bunch of cubes and see if you can somehow rearrange them in some other patterns. This is what's done here.

And yes, you could also try your method of guessing that there might be some kind of polynomial formula for this: $An^4+Bn^3+Cn^2+Dn+E$ .... though that is taking a bit of a risk: why would it be a polynomial? Why not some exponential function, for example? And even if it is a polynomial, why wouldn't it go beyond $n^4$?

Answer 2

A loose idea as to how to arrive at such formulas (albeit tedious, but most of the time works) is as follows. Let $f(x)$ be an integral polynomial of degree $d$. Let us say, we want to find a formula for $S(n)=f(0)+f(1)+\cdots +f(n)$. Knowing $\int x^d=x^{d+1}/d+1$, one guesses that the formula should be a polynomial of degree $d+1$. So write a polynomial $g(x)$ of degree $d+1$, whose coefficients are to be determined. These can be determined by writing a bunch of linear equations coming from $g(m)=S(m)$ for $0\leq m\leq d+1$. Once you have $g(x)$, attempt to prove the result using induction.

In your case, if you had guessed that the formula should be a quartic polynomial, the rest would have been easy.

Answer 3

If you know that $1^3 + \cdots + n^3$ is a $4$th degree polynomial $P(n)$ (for $n \geq 0$), you can write out $P(n) = a_4 n^4 + \cdots + a_1 n + a_0$ and solve the system of equations \begin{align*} P(0) &= 0 \\ \vdots& \\ P(4) &= 1^3 + \cdots + 4^3 \end{align*} to find the coefficients $a_i$. It's kind of messy (and maybe that's the method you used above), and it doesn't generalize easily to sums of higher powers $1^k + \cdots + n^k$. The punchline is Faulhaber's formula, which characterizes the coefficients in terms of the Bernoulli numbers (and, accordingly, requires a bit of machinery to prove).

So, why should you believe that such a polynomial $P$ exists? In essence, you're looking for a function with $P$ with \begin{align*} P(n) - P(n-1) = n^3 \end{align*} for $n \geq 1$. It's easy to note that if $P$ is a polynomial of degree $k > 0$, then $P(n) - P(n-1)$ is a polynomial of degree $k - 1$ in $n$. (The difference is clearly a polynomial of degree $\leq k$, and the coefficients of degree $k$ in $P(n)$ and $P(n-1)$ cancel.) To make this more precise, we could define an operator \begin{align*} (Sf)(n) &= \sum_{k=1}^n f(k) \end{align*} for reasonable functions $f$. Via, say, an inductive argument, we can prove that $Sf$ is a polynomial of degree $k$ for any nonzero polynomial of degree $k-1$; and, furthermore, \begin{align*} (Sf)(n) - (Sf)(n-1) = f(n). \end{align*} Thus we're effectively trying to compute $Sf$ for $f(n) = n^3$.

This sort of thing is called the finite-difference calculus, and there's a bunch of similar results on it. One particular thing I'll note is that it's often easier to work with the binomial coefficients \begin{align*} P_k(n) &= \binom{n}{k}, \end{align*} which is a polynomial of degree $k$. (More technically, there's a polynomial of degree $k$ that it equals for $n\geq 0$. These have the convenient property that \begin{align*} P_k(n) - P_k(n-1) = P_{k-1}(n-1), \end{align*} whereas (as you've seen if you've tried to do the computation by hand), the corresponding result for the polynomials $n^k$ are more complicated.

Answer 4

You can expand $$\left(\frac {n(n+1)}2\right)^2=\frac 14n^2(n^2+2n+1)=\frac 14n^4+\frac 12n^3+\frac 14n^2$$ and read off $A=\frac 14,B=\frac 12,C=\frac 14,D=0,E=0$ You can also solve simultaneous equations $$1^3=1=A1^4+B^3+C1^2+D1+E=A+B+C+D+E\\ 1^3+2^3=9=A2^4+B2^3+C2^2+D2+E=16A+8B+4C+2D+E$$ and three more similar. You should get the same result as above.

Answer 5

$0=E, 1=A+B+C+D+E, 1^3+2^3=9=16A+8B+4C+2D+E,$

$ 1^3+2^3+3^3=36=81A+27B+9C+3D+E, $

and $1^3+2^3+3^3+4^3=100=256A+64B+16C+4D+E \implies$

$ E=0, 7=14A+6B+2C, 33=78A+24B+6C, $ and $96=252A+60B+12C \implies$

$12=36A+6B$ and $54=168A+24B\implies$

$24A=6\implies A=\dfrac14$. Can you solve now for $B, C, $ and $D$?