# A question on proof by induction and cubic functions

the full question is

I equated each equation to f(n) and ended up with four equations and four unknowns. However I dont seem to be getting anywhere

Also does anyone know if this method will work for the series sum of r^3 ?

any help is hugely appreciated! thank you :)

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## 1 Answer

Presumably you got the following system of equations:

\left\{\begin{align*} &d=0\\ &a+b+c+d=1\\ &8a+4b+2c+d=5\\ &27a+9b+3c+d=14\;, \end{align*}\right.

which immediately reduces to

\left\{\begin{align*} &a+b+c=1\\ &8a+4b+2c=5\\ &27a+9b+3c=14\;. \end{align*}\right.

Eliminate $c$ by subtracting multiples of the first equation from the other two:

\left\{\begin{align*} &6a+2b=3\\ &24a+6b=11\;. \end{align*}\right.

Subtracting the second equation from $4$ times the first yields $2b=1$, so $b=\frac12$, and $6a+1=3$, or $a=\frac13$. Finally, from $a+b+c=1$ we have $c=\frac16$, and the cubic polynomial is

$$f(n)=\frac13n^3+\frac12n^2+\frac16n=\frac16\left(n\left(2n^2+3n+1\right)\right)=\frac16n(n+1)(2n+1)\;.$$

Now you have only to prove by induction that

$$\sum_{k=0}^nk^2=\frac16n(n+1)(2n+1)$$

for $n\ge 0$.

Yes, this approach will work to find a closed form for $\sum_{k=0}^nk^m$ for any positive integer $a$, though it’s pretty tedious if $a$ is much bigger than $2$; $m=3$ is quite feasible, however.

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Hi brian, thanks for that :) I think I made an error in my calculations because my second set of equations do not match yours. can i ask, if i were to come up with an algebraic expression in terms of n for the series 1^3 + 2^3+...+n^3 , and to prove by induction that it holds, the same method would work? –  Anona anon Dec 30 '12 at 10:29
Brian, correct me if I am wrong. You can form a $(4 \times 4)$-Vandermonde matrix using the initial four equations. Then there should be some closed formula for the inverse of this matrix that you can use to compute $(a,b,c,d)$. This method can be extended to the case of higher exponents. –  Haskell Curry Dec 30 '12 at 10:30
@Anonaanon: Yes, it would work; that’s what I was saying in the last sentence of my answer. –  Brian M. Scott Dec 30 '12 at 10:33
@Haskell: Something like that should work, though it’s not at all the way I think about it. (I tend to think of it in terms of finite calculus or, for the general case, exponential generating functions.) –  Brian M. Scott Dec 30 '12 at 10:43
@Anonaanon: No, it’s equal to $S(2)=\sum_{k=0}^2k^2=0^2+1^2+2^2=5$. –  Brian M. Scott Dec 30 '12 at 10:53
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