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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 )
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
Double induction example: $ 1 + q - Mathematics Stack Exchange Slightly relevant: you can see my answer on this thread for a proof that uses double induction (just to get you exposed to how the mechanics of a proof using double induction might work)
trigonometry - Why $\sin (n\pi) = 0$ and $\cos (n\pi)= (-1)^n . . . Since $2 \pi $ corresponds to a complete rotation, half a rotation will correspond to switching sign of both $\cos$ and $\sin$ (since it corresponds to a reflection through the origin) Since $\theta=0$ corresponds to $ (1,0)$ the result you desire follows