Formula for $1^2+2^2+3^2+. . . +n^2$ - Mathematics Stack Exchange $ (n+1)^3 - n^3 = 3n^2+3n+1$ - so it is clear that the $n^2$ terms can be added (with some lower-order terms attached) by adding the differences of cubes, giving a leading term in $n^3$ The factor 1 3 attached to the $n^3$ term is also obvious from this observation
Prove that $1^3 + 2^3 + . . . + n^3 = (1+ 2 + . . . + n)^2$ Do you know a simpler expression for $1+2+\ldots+k$? (Once you get the computational details worked out, you can arrange them more neatly than this; I wrote this specifically to suggest a way to proceed from where you got stuck )
linear algebra - How to tell if a set of vectors spans a space . . . Generically you don't know without examing the presumed "basis" vectors You do know that three vectors are sufficient (x,y,z) to span 3-space; any fourth vector must be a linear combination of (x,y,z) There is no more room