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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)
1 8, 1 4, 1 2, 3 4,7 8英寸分别是多少厘米? - 知乎 把1英寸分成8等分: 1 8 1 4 3 8 1 2 5 8 3 4 7 8 英寸。 This is an arithmetic sequence since there is a common difference between each term In this case, adding 18 to the previous term in the sequence gives the next term In other words, an=a1+d (n−1) Arithmetic Sequence: d=1 8
概率论与数理统计入门 - 知乎 反映了人们对犯第一类错误的最大容忍程度。 ② 置信水平1-α 是显著性水平的对立面，1-α是一个概率值，当α确定时，置信水平随之确定。 置信水平反映的是人们在面对第一类错误时，希望统计…