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 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?

share|improve this question
Please consider the special case n=4 and try to see where you are going wrong. –  Saugata 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
@user49521: What is special about n=4? –  Gineer Nov 18 '12 at 16:31
@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
show 7 more comments

5 Answers

up vote 7 down vote accepted

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

share|improve this answer
That's a beautiful trick! I think it's Bernouli's. –  Amihai Zivan Nov 19 '12 at 20:26
add comment

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 $n=1$.

share|improve this answer
add comment

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

share|improve this answer
add comment

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).

share|improve this answer
add comment

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 $f(0)=0$

share|improve this answer
add comment

Your Answer


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.