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Binomial expansion of $ (1-x)^n$ - Mathematics Stack Exchange I'm not sure how appropriate it is to answer questions this old, but compared to the methods above, I feel the easiest way to see the answer to this question is to take a = -x And substitute that into the binomial expansion: (1+a)^n This yields exactly the ordinary expansion Then, by substituting -x for a, we see that the solution is simply the ordinary binomial expansion with alternating
Mathematics Stack Exchange Q A for people studying math at any level and professionals in related fields
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