次方級數相加的問題(求公式)

如果m大於3推出來的結果很複雜，通常我們只列出2,3次的結果而已
我可以提供一種想法給大大參考，他能說明m次方一定有，但非常難算
(1+1)^3=1^3 +3*1*1^2 +3*1^2*1 +1^3
(1+2)^3=1^3 +3*1^2*2 +3*1*2^2 +2^3
(1+3)^3=1^3 +3*1^2*3 +3*1*3^2 +3^3
...........




(1+n)^3=1^3+ 3*1^2*n+ 3*1*n^2+ n^3

全部加起來，你會發覺前一向的右邊的n會與左邊的(n+1)相消，故和為:(1+ n)^3=n+ 3[n(n+ 1)/2] +3[所求]


整理一下，式子為3*[所求]=n(n +1)(2n+ 1)/2
故所求為n(n +1)(2n +1)/6

由這種方式我可以聯想到:如果要證明三次方，那就必須用到(1+k)^4
四次方就要用到(1+k)^5
如此可知，他一定可以推，但是結果會非常複雜

這只是其中一種想法，如果有更方便的方法的話可以提出來討論

We can use recrusive method to find out the answer.


Denote Sk (n)=1^k + 2^k + ... + n^k. There exist a recursive formula such that you can use the previous formula of Sk-1 (n), Sk-2 (n),... S1 (n) to find out Sk (n)
k+1C1Sk (n)+ k+1C2Sk-1 (n)+...+k+1Ck+1S0 (n)=(n+1)k+1-1

where S0 (n)=n.

For example, 2C1S1 (n)+2C2S0 (n)=(n+1)2-1

S1 (n)=(n2+2n-n)/2=n(n+1)/2
There is a similar formula which is related to Bernoulli numbers. However, since to find out those Bernoulli numbers also requires using other recursive formulae. So, I omit them here. On the other hand, some people have derived other recrusive formulae. Since those formulae are more awkward than the one I mentioned above, so I skip them here for simplicity.