$$\sum_{k=1}^n k^3=\frac{n^2 (n+1)^2}{4}$$
Please help
Asked
Active
Viewed 310 times
-1
-
2what have u tried and where r u stuck – Mr. Math Dec 09 '14 at 00:37
-
3@Guest For the future: you can flag the question, choosing "it is a duplicate" reason, to make other users aware of the duplicate. – Dec 09 '14 at 00:42
-
@Behaviour Thanks, I thought only high-power users could flag. – Guest Dec 09 '14 at 00:45
1 Answers
0
If $n=1$, it holds; Suppose that $n=m$, the equality also holds, i.e., $$\sum_{k=1}^mk^3=\frac{m^2(m+1)^2}{4}.$$
Now $$\sum_{k=1}^{m+1}k^3=\frac{m^2(m+1)^2}{4}+(m+1)^3=\frac{(m+1)^2(m+2)^2}{4}.$$
Paul
- 20,553
-
1The last equality hides the algebra which is the essence of the proof. I think it should be expanded showing all the steps. – marty cohen Dec 09 '14 at 02:11