Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

the full question is

enter image description here

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 :)

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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
2  
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
1  
@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
1  
@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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.