is there any proof for the sum of cubes except induction supposition? there are some proofs using induction in below page Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction
$\endgroup$ 02 Answers
$\begingroup$Write
$$(k+1)^4 = k^4+4k^3+6k^2+4k+1$$
or
$$(k+1)^4 -k^4 = 4k^3+6k^2+4k+1.$$
If you sum both sides from $k=1$ to $n$, the left side telescopes and you have
$$(n+1)^4 -1 = 4 \sum_{k=1}^{n} k^3 + 6\sum_{k=1}^{n} k^2 + 4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1.$$ Using known formulae for the last 3 sums gives
$$(n+1)^4 -1 = 4 \sum_{k=1}^{n} k^3 + 6 \frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}+n$$
and you can solve this equation for the remaining sum.
$\endgroup$ 3 $\begingroup$hint
begin by $$(n+1)^4=n^4+4n^3+6n^2+4n+1$$
... $$(3+1)^4=3^4+4.3^3+6.3^2+4.3+1$$ $$(2+1)^4=2^4+4.2^3+6.2^2+4.2+1$$ $$(1+1)^4=1^4+4.1^3+6.1^2+4.1+1$$
and sum to get
$$(n+1)^4=1+4S_3+6S_2+4S_1+n $$ with $$S_1=1+2+...+n=\frac {n (n+1)}{2}$$ $$S_2=1^2+2^2+...+n^2=\frac {n (n+1)(2n+1)}{6} $$
Now you can find your sum $S_3$.
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