Find the sum of the first $n$ terms of $\sum^n_{k=1}k^3$

Question

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?

Please consider the special case n=4 and try to see where you are going wrong. — Tatin, Nov 18 '12 at 16:21
This is not an arithmetic progression. The answer you are looking for is $\frac{n^2(n+1)^2}{4}$. — Gautam Shenoy, Nov 18 '12 at 16:24
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. — Gautam Shenoy, Nov 18 '12 at 16:31
@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? — Gineer, Nov 18 '12 at 16:36
The subject is redundant - the $\sum$ notation already includes the notion of the first $n$ terms. — Thomas Andrews, Nov 18 '12 at 16:38
@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. — Gautam Shenoy, Nov 18 '12 at 16:41
possible duplicate of How to calcuate sum of a "custom" series? — Norbert, Nov 18 '12 at 16:45
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. — Michael Hardy, Nov 18 '12 at 18:04
If it's still not clear, $1^3+4^3$ is not the same as $2^3+3^3$. — 2'5 9'2, Nov 19 '12 at 07:02

score 9 · Accepted Answer · edited Nov 18 '12 at 18:03

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 $$n^{4}+2n^{3}+n^{2}=n^{2}(n+1)^{2}=4S \implies \sum_{k=1}^{n}k^{3}=\frac{n^{2}(n+1)^{2}}{4} \square$$

That's a beautiful trick! I think it's Bernouli's. – Amihai Zivan Nov 19 '12 at 20:26 — Amihai Zivan, Nov 19 '12 at 20:26

score 4 · Answer 2 · answered Nov 19 '12 at 21:36

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 $n=1$.

score 2 · Answer 3 · answered Nov 18 '12 at 16:39

$$(k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$$ 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.

score 1 · Answer 4 · answered Nov 24 '12 at 12:18

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).

Souvik Dey · Answer 5 · 2013-10-17T13:25:56.707

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 $f(0)=0$

Find the sum of the first $n$ terms of $\sum^n_{k=1}k^3$

5 Answers5