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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?

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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
up vote 8 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 $$n^{4}+2n^{3}+n^{2}=n^{2}(n+1)^{2}=4S \implies \sum_{k=1}^{n}k^{3}=\frac{n^{2}(n+1)^{2}}{4} \square$$

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

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 $n=1$.

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

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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).

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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 $f(0)=0$

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