1*(2n-1)+3*(2n-3)+...+(2n-1)*1= Ans:1/3*n(2n² +1) 求運算過程,謝謝~~?

1(2n-1) + 3(2n-3) + ... + (2n-3)3 + (2n-1)1
= 1(2n-1) + 3(2n-3) + ... + (2n-3)(2n - (2n-3)) + (2n-1)(2n - (2n-1))
= (1 + 3 + ... + 2n-3 + 2n-1) 2n - (1² + 3² + ... + (2n-3)² + (2n-1)²)
= (1 + 2n-1)(n/2) 2n - (1² + 2² + 3² + ... + (2n-3)² + (2n-2)² + (2n-1)²) + (2² + 4² + 6² + ... + (2n-4)² + (2n-2)²)
= (1 + 2n-1)(n/2) 2n - (1² + 2² + 3² + ... + (2n-3)² + (2n-2)² + (2n-1)²) + 4(1² + 2² + 3² + ... + (n-2)² + (n-1)²)
= (1 + 2n-1)(n/2) 2n - (2n-1)(2n-1 + 1)(2(2n-1) + 1)/6 + 4(n-1)(n-1 + 1)(2(n-1) + 1)/6
= 2n³ - (2n-1)(2n)(4n-1)/6 + 4(n-1)(n)(2n-1)/6
= 2n(n² - (2n-1)(4n-1)/6 + (n-1)(2n-1)/3)
= ⅓ n(6n² - (2n-1)(4n-1) + 2(n-1)(2n-1))
= ⅓ n(6n² - (8n² - 6n + 1) + (4n² - 6n + 2))
= ⅓ n(2n² + 1)

令 1*(2n-1)+3*(2n-3)+...+(2n-1)*1 = Σ A(k)*B(k) , from k=1 to k=n
顯然 A(k) = 2k - 1

觀察 B(k) 之變化:
2n-1 , 2n-3 , ..... , 1
知 B(k) 為等差數列, 公差為 - 2,
故可設 B(k) = -2k + C , 再解出常數C:
B(1) = -2*1 + C = 2n - 1
C = 2n + 1 , 所以:
B(k) = -2k + 2n + 1

所求式
= Σ A(k)*B(k) , from k=1 to k=n
= Σ ( 2k - 1 )( -2k + 2n + 1 )
= Σ ( - 4k^2 + 4nk + 2k + 2k - 2n - 1 )
= Σ [ - 4k^2 + (4n+4)k - 2n - 1 ]
= -4 Σk^2 + (4n+4) Σ k + (-2n-1) Σ 1
= -4*n(n+1)(2n+1)/6 + (4n+4)*n(n+1)/2 + (-2n-1)*n
= (-2/3)(n^2+n)(2n+1) + (2n+2)(n^2+n) - 2n^2 - n
= (-2/3)(2n^3+3n^2+n) + (2n^3+4n^2+2n) - 2n^2 - n
= -(4/3)n^3 - 2n^2 - (2/3)n + 2n^3 + 4n^2 + 2n - 2n^2 - n
= (2/3)n^3 + (1/3)n
= (1/3)(2n^3+n)
= (n/3)(2n^2+1) ... Ans