講教2題數學題

方便起見，以下用 求和符號 ：

Σ 表示 Σ[k=1~n] ，即n個通項相加
例如 Σ(2k+1) = 3 + 5 + ... + (2n+1)

23(c)
考慮(a)中原式左邊
= Σ (2k-1)(2k+1)
= Σ (4k^2 -1)
= 4 (Σ k^2) - n
= (1/6) (2n-1)(2n+1)(2n+3) + (1/2)

所以
Σ k^2
= [(1/6) (2n-1)(2n+1)(2n+3) + (1/2) + n ] / 4
= (1/24) [(2n-1)(2n+1)(2n+3) + 3 + 6n ]
= (1/24) (8n^3 +12n^2+4n)
= (1/6) n(n+1)(2n+1)

即 1^2 + 2^2 +3^2 + ... + n^2 = (1/6) n(n+1)(2n+1)

24(b)

(a)部結果即 Σ(k/2) = (1/4) n (n+1)

考慮(b)中原式左邊
= Σ k(2k+2)
= Σ (2k^2 +2k)
= 2 (Σk^2 + Σk)
= 2 Σk^2 + 4Σ(k/2)
= 2 (1/6) n(n+1)(2n+1) + 4(1/4) n (n+1)
= (2/3) (n^3 +3n^2 +2n)
= (2/3) n(n+1)(n+3)

即 1*4+2*6+... + n(2n+2) = (2/3) n(n+1)(n+3)

(23c)

考慮 (2r + 1)(2r - 1) = 4r^2 - 1
=> r^2 = [(2r + 1)(2r - 1) + 1] / 4

所以 1^2 + 2^2 + 3^2 + ... n^2
= [(2(1) + 1)(2(1) - 1) + 1] / 4
+ [(2(2) + 1)(2(2) - 1) + 1] / 4
+ [(2(3) + 1)(2(3) - 1) + 1] / 4 + ...
+ [(2(n) + 1)(2(4) - 1) + 1] / 4
= {[(3 x 1) + 1] + [(5 x 3) + 1] + [(7 x 5) + 1] + ... + [(2n + 1)(2n - 1) + 1]} / 4
= [1 x 3 + 3 x 5 + 5 x 7 + ... + (2n - 1)(2n + 1) + n] / 4
= 代入後化簡

(24b)

1 x 4 + 2 x 6 + ... + n(2n + 2)
= 1 x (5 - 1) + 2 x (7 - 1) + ... + n(2n + 3 - 1)
= 1 x 5 - 1 + 2 x 7 - 2 + ... + n(2n + 3) - n
= [1 x 5 + 2 x 7 + ... + n(2n + 3)] - (1 + 2 + ... + n)
= [1 x 5 + 2 x 7 + ... + n(2n + 3)] - 2(1/2 + 2/2 + 3/2... + n/2)
= [1 x 5 + 2 x 7 + ... + n(2n + 3)] - 2(1/2 + 1 + 3/2... + n/2)
= 代入後化簡