證明題 (20分)

Now, denote the sign ''∑(n k=1)'' to ''∑''
∑ k = 1+2+3+...+n
=> 2∑k = (1+2+3+...+n) + (1+2+3+..+n)
=> 2∑k = (1+2+3+...+n) + ( n+(n-1)+...+1)
=> 2∑k = (1+n) + [2+(n-1)] + [3+(n-2)]+....+[n+1] <---有n個(n+1)
=> 2∑k = n(n+1)
=> ∑k = [n(n+1)]/2 -----------(1)
Then, we consider (k+1)^3 = k^3+3k^2+3k+1
=> (k+1)^3 - k^3 = 3k^2 + 3k +1
=> ∑ [(k+1)^3 - k^3 ] = ∑(3k^2+3k+1)
=> (n+1)^3 - 1^3 = 3∑(k^2) + 3∑(k) + ∑(1) ----(注4)
=> (n+1)^2 - 1 = 3∑(k^2) + 3[n(n+1)/2] + n (by (1))
=> ∑(k^2) = {(n+1)^2 - 1 -n -3[n(n+1)/2]}/3
=> ∑(k^2) = (1/6)[n(n+1)(2n+1)] --------(2)
Again, we consider (k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k +1
=> (k+1)^4 - k^4 = 4k^3 + 6k^2+ 4k +1
=> ∑ [(k+1)^4 - k^4] = ∑(4k^3+6k^2+4k+1]
=> (n+1)^4 -1^4 = 4∑k^3 + 6∑k^2 + 4∑k + ∑1
=> (n+1)^4 - 1 = 4∑k^3 + 6 [(1/6)(n)(n+1)(2n+1)]+4[n(n+1)/2]+n
(by (1),(2))
=> ∑k^3 = [n^2(n+1)^2]/4 -------(3)

注: (4). ∑(1) = ∑(n k=1) (1) = 1+1+...+1 =n
(5). ∑(n k=1) [ (k+1)^3 - k^3]
=[ ∑(n k=1) (k+1)^3 ] - [ ∑(n k=1) k^3]
=[ 2^3+3^3+...+(n+1)^3] - [ 1^3+2^3+3^3+...+n^3]
[中間個d項(3^3,4^3...,n^3)一加一減約左]
=(n+1)^3 - 1^3
留意..(1),(2),(3),(5) 公式都幾常用 [特別係(5)], 可多留意...
順便一題有關(3),
∑k^3 = [n^2(n+1)^2]/4 = [(n)(n+1)/2]^2 = [∑k]^2

http://hk.knowledge.yahoo.com/question/question?qid=7009110200126