2
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The question:
Find the sum of the first $n$ terms of $$\sum^n_{k=1}k^3$$ [Hint: consider $(k+1)^4-k^4$]
[Answer: $\frac{1}{4}n^2(n+1)^2$]

My solution:
$$\begin{align} \sum^n_{k=1}k^3&=1^3+2^3+3^3+4^3+\cdots+(n-1)^3+n^3\\ &=\frac{n}{2}[\text{first term} + \text{last term}]\\ &=\frac{n(1^3+n^3)}{2} \end{align}$$

What am I doing wrong?

  • 1
    Please consider the special case n=4 and try to see where you are going wrong.2012-11-18
  • 0
    This is not an arithmetic progression. The answer you are looking for is $\frac{n^2(n+1)^2}{4}$.2012-11-18
  • 1
    Hint: To get the sum, expand $(k+1)^4$. You will get some polynomial in RHS. Don't cancel the $k^4$ on RHS. Instead take summation k=1 to n on both sides, add 1 to both sides so that LHS becomes $\sum k^4 + (n+1)^4$ Then cancel the $\sum k^4$ on both sides. Now rearranget to get $\sum k^3$ in terms of $\sum k^2$, $\sum k$ and n.2012-11-18
  • 0
    @user49521: What is special about n=4?2012-11-18
  • 0
    @Gautam Shenoy: Why would I randomly try to expand $(k+1)^4$? I have an example in my book that I just found doing the same thing for $\sum_{k=1}^nk^2$ where they then expand $(k+1)^3$. Where does this come from?2012-11-18
  • 0
    The subject is redundant - the $\sum$ notation already includes the notion of the first $n$ terms.2012-11-18
  • 1
    @Gineer: The idea is in my answer. You go up a power. Look at my manipulation in the $(k+1)^4$ step. That's the idea.2012-11-18
  • 0
    possible duplicate of [How to calcuate sum of a "custom" series?](http://math.stackexchange.com/questions/115249/how-to-calcuate-sum-of-a-custom-series)2012-11-18
  • 0
    Here are two things you're doing wrong: (1) $(\text{first term}+\text{last term})/2$ works only if the sequence of terms is arithmetic. This one is not. (2) You're ignoring the hint.2012-11-18
  • 0
    If it's still not clear, $1^3+4^3$ is not the same as $2^3+3^3$.2012-11-19
  • 0
    [This thread](http://math.stackexchange.com/questions/61482/proving-the-identity-sum-limits-k-1n-k3-left-sum-limits-k-1n-k-ri) might be of interest, also.2012-11-19
  • 0
    http://en.wikipedia.org/wiki/Faulhaber's_formula2013-08-21

5 Answers 5

9

Let's take the suggested hint, and consider $$(k+1)^{4}-k^{4}=4k^{3}+6k^{2}+4k+1$$ Summing up both sides from $k=1$ to $n$. Notice that $$\sum_{k=1}^{n}[(k+1)^{4}-k^{4}]=[2^{4}-1^{4}]+[3^{4}-2^{4}]+\ldots+[n^{4}-(n-1)^{4}]+[(n+1)^{4}-n^{4}]$$
Cancelling, we get $(n+1)^{4}-1$. So altogether, $$(n+1)^{4}-1=4\sum_{k=1}^{n}k^{3}+6\sum_{k=1}^{n}k^{2}+4\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$$ Assuming you already know the results for $\sum1$, $\sum k$, $\sum k^{2}$, we can substitute them in: $$n^{4}+4n^{3}+6n^{2}+4n+1-1=4S+n(n+1)(2n+1)+2n(n+1)+n$$ $$[n^{4}+4n^{3}+6n^{2}+4n]-[2n^{3}+3n^{2}+n]-[2n^{2}+2n]-[n]=4S$$ Thus, simplifying further $$n^{4}+2n^{3}+n^{2}=n^{2}(n+1)^{2}=4S \implies \sum_{k=1}^{n}k^{3}=\frac{n^{2}(n+1)^{2}}{4} \square$$

  • 0
    That's a beautiful trick! I think it's Bernouli's.2012-11-19
4

Method 1

Using the binomial identity $$ \sum_{k=m}^{n-j}\binom{n-k}{j}\binom{k}{m}=\binom{n+1}{j+m+1}\tag{1} $$ with $j=0$ yields $$ \sum_{k=0}^n\binom{k}{m}=\binom{n+1}{m+1}\tag{2} $$ Using $(2)$ and the identity $$ k^3=6\binom{k}{3}+6\binom{k}{2}+\binom{k}{1}\tag{3} $$ we get that $$ \begin{align} \sum_{k=0}^nk^3 &=6\binom{n+1}{4}+6\binom{n+1}{3}+\binom{n+1}{2}\\ &=6\binom{n}{4}+12\binom{n}{3}+7\binom{n}{2}+\binom{n}{1}\\ &=\frac{n^2(n+1)^2}{4}\tag{4} \end{align} $$


Method 2

Using the Euler-Maclaurin Sum Formula, we get $$ \sum_{k=0}^nk^3=\frac14n^4+\frac12n^3+\frac14n^2+C\tag{5} $$ where we get $C=0$ by plugging in $n=1$.

2

$$(k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$$ Take sum from k=1 to n $$\sum_{k=1}^n(k+1)^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n $$ $$\sum_{k=2}^{n+1}k^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n $$ Add 1 on both sides

$$\sum_{k=1}^{n}k^4 + (n+1)^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n + 1$$ Cancel $\sum_{k=1}^{n}k^4$

$$ (n+1)^4 = \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n + 1$$

You need $\sum_{k=1}^n k^2, \sum_{k=1}^n k$ to solve it. Hope you can do the rest.

1

For a geometric solution, you can see theorem 3 on the last page of this PDF. Sorry I did not have time to type it here. This solution was published by Abu Bekr Mohammad ibn Alhusain Alkarachi in about A.D. 1010 (Exercise 40 of appendix E, page A38 of Stewart Calculus 5th edition).

0

Let $f(n)=\Big(\frac{n(n+1)}2\Big)^2$ , then$\space$$f(n-1)=\Big(\frac{n(n-1)}2\Big)^2$ ; now we know that

$(n+1)^2 - (n-1)^2=4n$ , this implies

$\space$ $n^2\Big((n+1)^2 - (n-1)^2\Big)=4n^3=$$\big(n(n+1)\big)^2 - \big(n(n-1)\big)^2$

$\implies$ $\frac{\big(n(n+1)\big)^2}4 - \frac{\big(n(n-1)\big)^2}4=$ $\Big(\frac{n(n+1)}2\Big)^2-\Big(\frac{n(n-1)}2\Big)^2=n^3$$=f(n)-f(n-1)$

$\implies$ $\sum_{n=1}^m \Big(f(n)-f(n-1)\Big) =$$\sum_{n=1}^mn^3$$=f(m)-f(0)=f(m)=\Big(\frac{m(m+1)}2\Big)^2$ , where we

have used $f(0)=0$